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IN  MEMORIAM 
FLORIAN  CAJORI 


Digitized  by  tine  Internet  Archive 

in  2007  with  funding  from 

IVIicrosoft  Corporation 


http://www.archive.org/details/dooleymathOOdoolrich 


VOCATIONAL 
MATHEMATICS 


BY 


WILLIAM   H.   DOOLEY 

AUTHOR    OP    "  VOCATIONAL    MATHEMATICS    FOB 
GIRLS,"    "textiles,"    ETC. 


REVISED 


D.    C.    HEATH   &   CO.,   PUBLISHERS 
BOSTON  NEW  YORK  CHICAGO 


Copyright,  191 5, 
By  D.  C.  Heath  &  Co. 

2  B  O 

CAJORI 


PREFACE 

The  author  has  had,  during  the  last  ten  years,  considerable 
experience  in  organizing  and  conducting  intermediate  and  sec- 
ondary technical  schools.  During  this  time  he  has  noticed  the 
inability  of  the  regular  teachers  in  mathematics  to  give  the 
pupils  the  training  in  commercial  and  rule  of  thumb  methods 
of  solving  mathematical  problems  that  are  so  necessary  in 
everyday  life.  A  pupil  graduates  from  the  course  in  mathe- 
matics without  being  able  to  "  commercialize  "  or  apply  his 
mathematical  knowledge  in  such  a  way  as  to  meet  the  needs 
of  trade  and  industry. 

It  is  to  overcome  this  difficulty  that  the  author  has  prepared 
this  book  on  vocational  mathematics.  He  does  not  believe  in 
doing  away  with  the  regular  course  of  mathematics  but  in 
supplementing  it  with  a  practical  course.  This  course  may  take 
the  place  of  the  first  year  algebra  and  the  first  year  geometry 
in  vocational  classes  in  which  it  is  not  desirable  to  give  the 
traditional  course  in  algebra  and  geometry. 

This  book  may  be  used  by  the  regular  teacher  in  mathe- 
matics and  by  the  shop  teacher.  It  can  be  used  in  the  shop 
in  teaching  mathematics  and  in  providing  drill  problems  upon 
the  shop  work.  A  course  based  upon  the  contents  of  the  book 
should  be  provided  before  pupils  finish  their  training,  so  that 
they  may  become  skillful  in  applying  the  principles  of  mathe- 
matics to  the  daily  needs  of  manufacturing  life. 

In  revising  the  manuscript  the  author  has  had  the  assistance 
of  his  teachers  in  the  Lawrence  Industrial  School,  the  Lowell 
Industrial  School,  the  Fall  River  Technical  High  School,  and 


9i6287 


IV  PREFACE 

of  many  other  teachers,  practical  men,  and  manufacturing  firms. 
Valuable  material  has  also  been  obtained  from  standard  hand- 
books, such  as  Kent's. 

To  mention  the  names  of  all  persons  to  whom  the  author  is 
indebted  is  impossible.  Acknowledgment  should  be  made  to 
the  following  persons  and  firms  who  have  kindly  consented 
to  furnish  cuts  and  information,  and  have  offered  valuable 
suggestions  and  problems :  Mr.  George  W.  Evans,  Principal 
Charlestown  (Mass.)  High  School;  Dr.  David  Snedden; 
Mr.  Charles  E.  Allen ;  Mr.  Fred.  W.  Turner ;  Mr.  Peter  Gart- 
land.  Principal  South  Boston  High  School ;  Mr.  John  Casey, 
Worcester  (Mass.)  Trade  School ;  Mr.  John  L.  Sullivan,  Princi- 
pal Chicopee  (Mass.)  Industrial  School;  Mr.  William  Hunter, 
Fitchburg  Industrial  Course;  Mr.  J.  Gould  Spofford,  Princi- 
pal Quincy  Industrial  School ;  Mr.  Edward  K.  Markham,  Cam- 
bridge Technical  High  School;  Principal  Joseph  J.  Eaton, 
Saunders  Trades  School,  Yonkers,  N,  Y. ;  Miss  Bessie  King- 
man, Brockton  (Mass.)  High  School ;  Mr.  E.  W.  Boshart, 
Director  of  Industrial  Arts,  Mt.  Vernon,  N.  Y. ;  Mr.  G.  A. 
Boate,  Technical  High  School,  Newton,  Mass. ;  Mr.  G.  R. 
Smith,  Bradford,  England ;  Mr,  H.  R.  Carter,  Belfast,  Ireland ; 
New  York  Central  Lines,  Apprenticeship  Department ;  Mr.  H. 
E.  Thomas,  Tuskegee  Institute,  Ala. ;  Brown  &  Sharp  Co. ; 
Simond's  Guide  for  Carpenters ;  R.  M.  Starbuck  &  Sons ; 
Garvin  Machine  Co. ;  The  Crane  Co. ;  The  William  Powell  Co. ; 
Crosby  Steam  Gage  and  Valve  Co. ;  Mr.  Peter  Lobben ;  Mr.  H. 
P.  Faxon ;  Bardons  and  Oliver ;  Stanley  Rule  and  Level  Co. ; 
Pittsburgh  Valve  Foundry  and  Construction  Co. ;  Becker 
Milling  Machine;  Braeburn  Steel  Co.;  Fore  River  Ship  Build- 
ing Co. ;  Engineering  Workshop  Machines  and  Processes  ; 
American  Injector  Co. ;  B.  F.  Sturtevaut  Co. ;  E.  W.  Bliss  Co. ; 
Dillon  Boiler  Works;  Whitcomb-Blaisdell  Co.;  Hoefer  Drill 
Co. ;  Bradford  Lathe  Co. ;  and  Detroit  Screw  Works. 

The  author  will  be  very  thankful  for  any  suggestions  relating 
to  the  work, 


CONTENTS 


PART   I  — REVIEW  OF   ARITHMETIC 

OUXPTXR  PAUS 

I.     Essentials  of  Arithmetic 1 

Fundamental  Processes.  Fractions.  Decimals.  Com- 
pound Numbers.  Percentage.  Ratio  and  Proportion. 
Involution.     Evolution. 

II.    Mensuration 62 

Circle.  Triangles.  Quadrilaterals.  Polygons.  El- 
lipse.    Pyramid.     Cone.     Sphere.     Similar  Figures. 

III.    Interpretation  of  Results 78 

Reading  of  Blue  Print.  Drawing  to  Scale.  Methods 
of  Solving  Examples. 


PART  II  — CARPENTERING   AND  BUILDING 

IV.     Woodworking 83 

Board  Measure.     Short  Methods.    Tables. 
V.    Construction 89 

Frame  and  Roof.     Lathing. 
VI.    Building  Materials 94 

Mortar.     Plaster.     Bricks.     Stone  Work.     Cement. 

Shingles.     Slate.     Clapboards.     Flooring.     Stairs. 

PART  III  — SHEET  METAL  WORK 

Vn.    Die  Cutting 107 

Blanking  Dies.    Combination  Dies.    Standard  Gauge. 
Tables. 

V 


VI  CONTENTS 


PART  IV  — BOLTS,   SCREWS,   AND   RIVETS 

CHAPTER  PAGE 

VIII.     Bolts,  Screws,  and  Rivets 126 

Kinds  of  Bolts.  Rivets.  Nails.  Tacks.  Kinds  of 
Screws.  Screw  Threads.  V-Shaped.  U.  S.  Stand- 
ard.    Acme  Standard.     Square.     Micrometer. 


PART  V  — SHAFTS,  PULLEYS,  AND  GEARS 

IX.     Shafts  and  Pulleys  . 146 

Belting.     Arc  of  Contact.     Speed.     Countershafts. 

X.     Gearing 159 

Pitch.     Trains  of  Gears. 

PART   VI  — PLUMBING   AND   HYDRAULICS 

XL     Plumbing  and  Hydraulics        .        .        .        .        .     172 
Tanks.      Drainage   Pipes.      Weight  of  Lead   Pipe. 
Capacity   of  Pipe.      Atmospheric   Pressure.     Water 
Pressure.    Water  Traps.    Velocity  of  Water.    Water 
Power     Density  of  Water.     Specific  Gravity. 

PART   VII  — STEAM   ENGINEERING 

XII.     Heat 195 

Heat  Units.  Temperature.  Value  of  Coal.  Intrinsic 
Heat.  Latent  Heat.  Value  of  Water  and  Steam. 
Steam  Heating. 

XIIL    Boilers 203 

Kinds  of  Boilers.  Tensile  Strength.  Safe  Working 
Pressure.  Boiler  Tubes.  Safety  Valves.  Super- 
heated Steam.    Boiler  Pumps. 

XIV.     Engines 222 

Kinds  of  Engines.  Fly  Wheel.  Horse  Power.  Di 
ameter  of  Cylinder.  Diameter  of  Supply  Pipe.  Steam 
Indicator. 


CONTENTS  Vll 


PART   VITI.     ELECTRICAL   WORK 

caAPTBB  PA«B 

XV.     CoMMKKriAi.  Elkctricity 231 

Amperes.  Volts.  Ohms.  Ammeter.  Voltmeter. 
Series  and  Parallel  Circuits.  Power  Measurement. 
Resistance.     Size  of  Wire. 


PART   IX  — MATHEMATICS   FOR  MACHINISTS 

XVI.     Matkkials 251 

Cast  Iron.  Wrought  Iron.  Steel.  Copper.  Brass. 
Weight  of  Steel. 

XVn.     Lathes 255 

Engine  Lathe,  (iear  and  Pitch.  Adjusting  Gears. 
Compound  Lathe.  To  cut  Double  or  Multiple 
Threads.  Machine  Speeds.  Net  Power  for  Cutting 
Iron  or  Steel.    Surface  Speeds.' 

XVIII.     Planers,  Shapers,  and  Drilling  Machines  .    268 

Planer.     Planer  and  Shaper.     Drilling  Machine. 

PART  X  — TEXTILE   CALCULATIONS 

XIX.     Yarns 276 

Materials.     Kinds.     Weight. 

APPENDIX 

Metric  System 291 

Graphs 296 

Formulas 298 

Logarithms 315 

Trigonometry 320 

Tables  of  Formulas 330 

Index 337 


NOTE   TO   TEACHERS 

The  author  has  found  that  it  adds  to  the  interest  of  the 
subject  if  the  teacher  will  provide  models,  instruments,  etc., 
in  the  class  and  explain  the  subject  matter  and  problems  in 
terms  of  the  actual  subject. 

If  the  pupils  have  not  had  a  course  in  Algebra,  it  is  better 
to  take  up  the  subject  of  Formulas  in  the  Appendix  before 
beginning  the  subject  of  Mensuration. 


viU 


VOCATIOIN^AL  MATHEMATICS 

PART   I  — REVIEW   OF   ARITHMETIC 

CHAPTER    I 

Notation  and  Numeration 

A  unit  is  one  thing ;  as,  one  book,  one  pencil,  one  inch. 
A  number  is  made  up  of  units  and  teUs  how  many  units  are 
taken. 

An  integer  is  a  whole  number. 

A  single  figure  expresses  a  certain  number  of  units  and  is  said  to  be  in 
the  units  column.  For  example,  5  or  8  is  a  single  figure  in  the  units 
column  ;  63  is  a  number  of  two  figures  and  has  the  figure  3  in  the  units 
column  and  the  figure  5  in  the  tens  column,  for  the  second  figure 
represents  a  certain  number  of  tens.  Each  column  has  its  own  name, 
as  shown  below. 


1       I       S       1 


3     8,     6      9      6,     4     O      7,     1      2      5 


Reading  Numbers.  —  For  convenience  in  reading  and  writing 
numbers  they  are  separated  into  groups  of  three  figures  each 
by  commas,  beginning  at  the  right  : 

138,695,407,125. 
The  first  group  is  125  units. 
The  second  group  is  407  thousands. 
The  third  group  is  695  millions. 
The  fourth  group  is  138  billions. 
1 


VOfSATjLONAL  MATHEMATICS 


'  ^^e;  ^rec^ding  fiui»be,r  is  read  one  hundred  thirty-eight 
billion,  six  hundred  ninety-five  million,  four  hundred  seven 
thousand,  one  hundred  twenty-five ;  or  138  billion,  695  million, 
407  thousand,  125. 

Standard  Mathematics  Sheet.  —  To  avoid  errors  in  solving  problems 
the  work  should  be  done  in  such  a  way  as  to  show  each  step,  and  should 
make  it  easy  to  check  the  answer  when  found.  A  sheet  of  paper  of 
standard  size,  8|  in.  by  11  in.,  should  be  used.  Rule  this  sheet  as  in  the 
following  diagram,  set  down  each  example  with  its  proper  number  in  the 
margin,  and  clearly  show  the  different  steps  required  for  the  solution. 
To  show  that  the  answer  obtained  is  correct,  the  proof  should  follow  the 
example  itself. 

Standard  Mathematics  Sheet 

8h  in. 


1. 

t 
It 

John  Smith  — 100 
Vocational  Arithmetic 

10-2-m    No.  10 

1,203                      20 
2,672                      2^ 
31,118                     23 
480                      10 
39 
19,883 
55,395    Ans. 

2. 

Proof: 

3. 

The  pupil  should  write  or  print  his  name  and  class,  the  date  when  the 
problem  is  finished,  and  the  number  of  the  problem  on  the  Standard 


REVIEW  OP  ARITHMETIC  3 

Mathematics  Sheet.  If  the  question  contains  several  divisions  or  prob- 
lems, they  should  be  tabulated  —  (a),  (6),  etc.  — at  the  left  of  the  prob- 
lems inside  the  margin  line.  Tiiere  should  be  a  line  drawn  between 
problems  to  separate  them. 

Addition 

Addition  is  the  process  of  finding  the  sum  of  two  or  more 
numbers.  The  result  obtained  by  this  process  is  called,  the 
sum  or  amount. 

The  sign  of  addition  is  an  upright  cross,  +,  called  plus.  The 
sign  is  placed  between  the  two  numbers  to  be  added. 

Thus,  9  inches  +  7  inches  (read  nine  inches  plus  seven  inches). 

The  sign  of  equality  is  two  short  horizontal  parallel  lines,  =, 
and  means  equals  or  equal  to. 

Thus,  the  statement  that  8  feet  +  6  feet  =  14  feet,  means  that  six  feet 
added  to  eight  feet  (or  8  feet  plus  6  feet)  equals  fourteen  feet. 

To  find  the  sum  or  amount  of  two  or  more  numbers. 

Example.  —  What  is  the  total  weight  of  a  machine  made  up 
of  parts  weighing  1203,  2672,  31,118,  480,  39,  and  19,883  lb., 
respectively  ? 

[The  abbreviation  for  pounds  is  lb.] 

The  sum  of  the  units  column  is  3  +  9  -f  0 
-1-8  +  2  +  3  =  26  units  or  20  and  5  more  ; 
20  is  tens,  so  leave  the  6  under  the  units 
column  and  add  the  2  tens  in  the  tens  column. 
The  sura  of  the  tens  column  is  2  +  8  +  3  +  8 
+  1  +  7  +  0  =  29  tens.  29  tens  equal  2  hun- 
dreds and  9  tens.  Place  the  9  tens  under 
the  tens  column  and  add  the  2  hundreds  to 
the  hundreds  column,  2  +  8  +  4  +  1+6 
+  2  =  23  hundreds ;  23  hundreds  are  equal  to  2  thousands  and  3  hundreds. 
Place  the  3  hundreds  under  the  hundreds  column  and  add  the  2  thousands 
to  the  next  column.  2  +  9  +  1+2  +  1  =  15  thousands  or  1  ten-thousand 
and  6  thousands.    Add  the  1  ten-thousand  to  the  ten-thousands  column 


1,203 

2p 

2,672 

29 

31,118 

2;^ 

480 

Ip 

39 

19,883 

55,395  lb. 

4  VOCATIONAL  MATHEMATICS 

and  the  sum  isl+l-)-3  =  5.     Write  the  5  in  the  ten-thousands  column. 
Hence,  the  sum  or  weight  is  55,395  lb. 

Proof.  —  Repeat  the  process,  beginning  at  the  top  of  the  right-hand 
column. 

Exactness  is  very  important  in  arithmetic.  There  is  only- 
one  correct  answer.  Therefore  it  is  necessary  to  be  accurate 
in  performing  the  numerical  calculations.  A  check  of  some 
kind  should  be  made  on  the  work.  The  simplest  check  is  to 
estimate  the  answer  before  solving  the  problem.  A  comparison 
can  be  made  between  them.  If  there  is  a  great  discrepancy 
then  the  work  is  probably  wrong.  It  is  also  necessary  to  be 
exact  in  reading  the  problem. 

EXAMPLES 

1.  Write  the  following  numbers  as  figures  and  add  them : 
Seventy-five  thousand  three  hundred  eight ;  seven  million  two 
hundred  five  thousand  eight  hundred  forty -nine. 

2.  In  a  certain  year  the  total  output  of  copper  from  the 
mines  was  worth  $58,638,277.86.  Express  this  amount  in 
words. 

3.  Solve  the  following: 

386  -h  5289  -f  53666  +  3001  -f  291  +  38  =  ? 

4.  On  a  shelf  there  were  three  kegs  of  bolts.  The  first  keg 
weighed  203  lb.,  the  second  171  lb.,  and  the  third  93  lb.  How 
many  pounds  of  bolts  were  there  on  the  shelf  ? 

5.  On  the  platform  in  an  electrical  shop  there  were  a  motor- 
generator  and  two  motors.  The  motor-generator  weighed  275 
lb.,  one  of  the  motors  385  lb.,  and  the  other  motor  492  lb. 
What  weight  did  the  platform  support  ? 

6.  Solve  the  following : 

6027  +  836  -f  4901  +  3,800,031  -f  28,639  +  389,661  =  ? 

7.  Four  coal  sheds  in  a  shop  contained  respectively  1498 
lb.,  4628  lb.,  6125  lb.,  and  12,133  lb.  What  was  the  whole 
amount  of  coal  in  the  shop  ? 


REVIEW  OF  ARITHMETIC  5 

8.  An  engineer  ordered  coal  and  found  that  the  first  load 
weighed  5685  lb.,  the  second  5916  lb.,  the  third  6495  lb.,  and 
the  fourth  5280  lb.     What  was  the  total  weight  ? 

9.  Wire  for  electric  lights  was  run  around  four  sides  of 
three  rooms.  If  the  first  room  was  13  ft.  long  and  9  ft.  wide ; 
the  second  18  ft.  long  and  18  ft.  wide ;  and  the  third  12  ft.  long 
and  7  ft.  wide,  what  was  the  total  length  of  wire  required? 
Remember  that  electric  lights  require  two  wires. 

10.  Find  the  sum : 

46  lb  +  135  lb.  +  72  lb.  +  39  lb.  +  427  lb.  +  64  lb. 
-h  139  lb. 

Subtraction 

Subtraction  is  the  process  of  finding  the  difference  between 
two  numbers,  or  of  finding  what  number  must  be  added  to  a 
given  number  to  equal  a  given  sum.  The  minuend  is  the  num- 
ber from  which  we  subtract ;  the  subtrahend  is  the  number  sub- 
tracted; and  the  difference  or  remainder  is  the  result  of  the 
subtraction. 

The  sign  of  subtraction  is  a  short  horizontal  line,  — ,  called 
minuSf  and  is  placed  before  the  number  to  be  subtracted. 

Thus,  12  —  8  =  4  is  read  twelve  minus  (or  less)  eight  equals  four. 

To  find  the  difference  of  two  numbers. 

Example.  —  A  reel  of  wire  contained  8074'.     If  4869'  were 
used  in  wiring  a  house,  how  many  feet  remained  on  the  reel  ? 
[Feet  and  mches  are  represented  by  '  and  "  respectively.] 

Minuend  8074  ft.  Write  the  smaller  number  under  the 

Subtrahend      4869  greater,  with  units  of  the  same  order  in 

Pfmn  'ndpr       S^H)5  ft        ^^^  same  vertical  line.     9  cannot  be  taken 

from  4,  so  change  1  ten  to  units.    The  1  ten 

that  was  changed  from  the  7  tens  makes  10  units,  which  added  to  the  4 

units  makes  14  units.    Take  9  from  the  14  units  and  6  units  remain. 

Write  the  5  under  the  unit  column.    Since  1  ten  was  changed  from  7  tens, 

there  are  6  tens  left,  and  6  from  0  leaves  0.     Write  0  under  the  tens  col- 

unm.    Next,  8  hundred  cannot  be  taken  from  0  hundred,  so  1  thousand 


6  VOCATIONAL  MATHEMATICS 

(ten  hundred)  is  changed  from  the  thousands  column.  8  hundred  from 
10  hundred  leave  2  hundred.  Write  the  2  under  the  hundreds  column. 
Since  1  thousand  has  been  taken  from  the  8  thousand,  there  are  left  7 
thousand  to  subtract  the  4  thousand  from,  which  leaves  3  thousand. 
Write  3  under  the  thousand  column.     The  whole  remainder  is  3205  ft. 

Proof.  —  If  the  sum  of  the  subtrahend  and  the  remainder  equals  the 
minuend,  the  answer  is  correct, 

EXAMPLES 

1.  Subtract  1001  from  79,999. 

2.  A  box  contained  one  gross  (144)  of  wood  screws.  If  48 
screws  were  used  on  a  job,  how  many  screws  were  left  in 
the  box  ? 

3.  What  number  must  be  added  to  3001  to  produce  a  sum 
of  98,322? 

4.  A  machinist  had  castings  of  different  kinds  on  hand  in 
the  morning,  weighing  6018  lb.  He  used  during  the  day  911 
lb.  of  castings.     How  many  pounds  were  left  ? 

5.  The  value  of  electrical  equipment  produced  by  one  firm 
in  a  certain  year  was  $  102,000,000,  and  in  the  previous  year  it 
was  $93,000,000.  What  was  the  difference  between  the  two 
years  ? 

6.  In  the  coal  shed  of  a  factory  there  were  52,621  lb.  of 
coal  at  the  beginning  of  the  week.  Monday  6122  lb.  were 
used ;  Tuesday  5928  lb  ;  Wednesday  2448  lb.  were  received 
and  5831  lb.  used,  (a)  How  much  coal  was  used  during  those 
days  ?  (6)  How  many  pounds  of  coal  were  there  in  the  shed 
on  Thursday  morning  ? 

7.  Monday  morning  an  engineer  had  72  gallons  of  cylinder 
oil.  Monday  he  used  8  gallons,  Tuesday  12  gallons,  and 
Wednesday  9  gallons,  (a)  How  many  gallons  did  he  use  ? 
(b)  How  many  gallons  of  oil  did  he  have  on  hand  Thursday 
morning  ? 

a   69,221-3008  =  ? 


REVIEW  OF  ARITHMETIC  7 

9.  A  reel  contained  13,211'  of  wire;  1112'  were  used  in 
wiring  a  house  and  341'  in  wiring  another  house,  (a)  How 
many  feet  of  wire  were  used  for  the  houses  ?  (6)  How  many 
feet  of  wire  were  left  on  the  reel  ? 

10.  Several  castings  were  placed  on  a  platform.  Their  total 
weight  was  1625  lb.  One  casting  weighing  215  lb.  and  an- 
other weighing  75  lb.  were  taken  away,  (a)  What  was  tlie 
weight  of  the  castings  taken  away  ?  (b)  What  was  the  weight 
of  the  remaining  castings  ? 

Multiplication 

Multiplication  is  the  process  of  finding  the  product  of  two 
numbers.  The  numbers  multiplied  together  are  called  factors. 
The  vudtipUcand  is  the  number  multiplied ;  the  multiplier  is  tlie 
number  multiplied  by;  and  the  result  is  called  the  product. 

The  sign  of  multiplication  is  an  oblique  cross,  x ,  which  means 
midtiplied  by  or  times. 

Thus,  8x3  may  be  read  8  multiplied  by  3,  or  8  times  3. 

To  find  the  product  of  two  numbers. 

Example.  —  An  iron  beam  weighed  24  lb.  per  foot  of  length. 
What  was  the  weight  of  a  beam  17  feet  long  ? 

Midtiplicand      24  lb.  Write  the  multiplier  under  the  multipli- 

jif  jf  J-  -I-  cand,  units  under  units,  tens  under  tens, 

^  — —  etc.     7  times  4  units  equal  28  units,  which 

^^^  are  2  tens  and  8  units.    Place  the  8  under 

^4  the  units  column.     The  2  tens  are  to  be 

Product  408  lb.       added  to  the  tens  product.    7  times  2  tens 

are  14  tens  -j-  the  2  tens  are  16  tens,  or  1 
hundred  and  6  tens.  Place  the  6  tens  in  the  tens  column  and  the  1  hun- 
dred in  the  hundreds  column.  168  is  a  partial  product.  To  multiply  by 
the  1,  proceed  as  before,  but  as  1  is  a  ten,  write  the  first  number,  which 
is  4  of  this  partial  product,  under  the  tens  column,  and  the  next  number 
under  the  hundreds  column,  and  so  on.  Add  the  partial  products  and 
their  sum  is  the  whole  product,  or  408  lb. 


8  VOCATIONAL  MATHEMATICS 

EXAMPLES 

1.  An  electrical  job  required  the  following  labor :  5  men,  42 
hours  each ;  6  men,  4  hours  each ;  12  men,  14  hours  each ;  7 
men,  12  hours  each ;  and  3  men,  6  hours  each.  Find  the  total 
number  of  hours  time  on  the  job. 

2.  Multiply  839  by  291. 

3.  A  machinist  sent  in  the  following  order  for  bolts :  12 
bolts,  6  lb.  each;  9  bolts,  7  lb.  each;  11  bolts,  3  lb.  each;  6 
bolts,  2  lb.  each;  and  20  bolts,  3  lb.  each.  What  was  the 
total  weight  of  the  order  ? 

4.  Find  the  product  of  1683  and  809. 

'    To  multiply  by  10,  100,  1000,  etc.y  annex  as  many  ciphers  to 
the  multiplicand  as  there  are  ciphers  iii  the  multiplier. 
Example.  — 864  x  100  =  86,400. 

EXAMPLES 

Multiply  and  read  the  answers  to  the  following : 

1.  869  X  10  a  100  X  500 

2.  1011  X  100  9.  1000  X  900 

3.  10,389  X  1000  10.  10,000  x  500 

4.  11,298  X  30,000  11.  10,000  x  6000 

5.  58,999  X  400  12.  1,000,000  x  6000 

6.  681,719  X  10  13.  1,891,717  x  400 

7.  801,369  x  100  14.  10,000,059  x  78,911 

Division 

Division  is  the  process  of  finding  how  many  times  one  number 
is  contained  in  another.  The  dividend  is  the  number  to  be 
divided ;  the  divisor  is  the  number  by  which  the  dividend  is 
divided;  the  quotient  is  the  result  of  the  division.  When  a 
number  is  not  contained  an  equal  number  of  times  in  another 
number  what  is  left  over  is  called  a  remainder. 


REVIEW  OF  ARITHMETIC  9 

The  sign  of  division  is  -j-,  and  wlieii  placed  between  two 
numbers  signifies  tliat  the  first  is  to  be  divided  by  the  second. 

Thus,  66  -!-  8  is  read  56  divided  by  8. 

Division  is  also  indicated  by  writing  the  dividend  above  the 
divisor  with  a  line  between. 

Thus,  *^ ;  this  is  read  56  divided  by  8. 

In  division  we  are  given  a  product  and  one  of  the  factors  to 
find  the  other  factor. 

To  find  how  many  times  one  number  is  contained  in  another. 

Example.  —  A  machinist  had  8035  rivets  which  he  wished  to 
arrange  in  groups  of  three.  How  many  such  groups  did  he 
have  ?     How  many  rivets  did  he  have  left  over  ? 

Quotient  Place  the  numbers  in  the  manner  in- 

2678  dicated  at  the  left.    8  thousand  is  in  the 

Divisor  3)8035  Dividend       thousands  column.    The  nearest  8  thou- 
g  sand  can  be  divided  into  groups  of  3  is  2 

OA  (thousand)  times,  which  gives  (>  thousand. 

^^  Write  2  as  the  first  figure  in  tlie  quotient 

over  8  in  the  dividend.     Place  the  6  (thou- 

•^"^  sand)  under  the  8  thousand  and  subtract ; 

21  the  remainder  is  2  thousand  or  20  hundred. 

25  8  is  contained  in  20  himdred  6  hundred 

24  times  or  18  hundred  and  2  hundred  re- 

R  m  ind       ~T  mainder.     Write  6  as  the  next  figure  in 

the  quotient.  Add  the  3  tens  in  the  divi- 
dend to  the  2  hundred  or  20  tens,  and  23  tens  is  the  next  dividend  to  be 
divided.  3  is  contained  in  23  tens  7  times  or  21  tens  with  a  remainder 
of  2  tens.  Write  7  as  the  next  figure  in  the  quotient.  2  tens  or  20  units 
plus  the  6  units  from  the  quotient  make  25  units.  3  is  contained  in  26, 
8  times.  Write  8  as  the  next  figure  in  the  quotient.  24  units  subtracted 
from  25  units  leave  a  remainder  of  1  unit.  Then  the  answer  is  2678 
groups  of  rivets  and  1  rivet  left  over. 

Proof.  —  Find  the  product  of  the  divisor  and  quotient,  add  the  re- 
mainder, if  any,  and  if  the  sum  equal  the  dividend,  the  answer  is  correct. 


10  VOCATIONAL  MATHEMATICS 

EXAMPLES 

1.  A  strip  of  plate  measures  85"  in  length.  How  many 
pieces  6"  long  can  be  cut  from  it?  AVould  there  be  a  re- 
mainder ? 

2.  How  many  pieces  2"  long  can  be  cut  from  a  brass  plate 
62'  long,  if  no  allowance  is  made  for  waste  in  cutting  ? 

3.  If  the  cost  of  constructing  362  miles  of  railway  was 
$4,561,200,  what  was  the  cost  per  mile  ? 

4.  If  a  job  which  took  379  hours  was  divided  equally 
among  25  men,  how  many  even  hours  would  each  man  w^ork, 
and  how  much  overtime  would  one  of  the  number  haye  to  put 
in  to  complete  the  job  ? 

5.  The  "  over-all  "  dimension  on  a  drawing  was  18'  9". 
The  distance  was  to  be  spaced  off  into  14-inch  lengths,  beginning 
at  one  end.  How  many  such  lengths  could  be  spaced  ?  How 
many  inches  would  be  left  at  the  other  end  ? 

6.  If  a  locomotive  consumed  18  gallons  of  fuel  oil  per  mile 
of  freight  service,  how  far  could  it  run  with  2036  gallons  of  oil  ? 

7.  If  48  screws  w^eigh  one  pound,  how  many  cases  each 
containing  36  screws  could  be  filled  from  a  stock  of  29  lb.  of 
screws  ? 

8.  A  plate  measures  68"  in  length.  How  many  pieces  15" 
long  can  be  cut  from  it ?     How  much  will  be  left  over? 

9.  A  machinist  desired  to  know  the  number  of  lots  of  42 
lb.  each,  contained  in  3276  lb.  of  screws.     Solve. 

10.  If  a  freight  train  made  a  distance  of  112  miles  in  8 
hours,  what  was  the  average  speed  per  hour  ? 

11.  If  the  circumference  of  the  driving  wheel  of  a  loco- 
motive was  22  feet,  how  many  turns  did  it  make  in  going  88 
miles  ? 

12.  Divide  38,910  by  3896. 


REVIEW   OF  ARITHMETIC  11 

REVIEW   EXAMPLES 

1.  A  load  of  castings  came  to  a  machine  shop.  It  was  not 
desirable  to  weigh  the  castings  on  the  wagon,  so  they  were 
weighed  in  6  lots  as  follows:  196  lb.,  389  lb.,  876  lb.,  899  lb., 
212  lb.,  and  847  lb.     What  was  the  total  weight? 

2.  Five  steel  bars  are  placed  end  to  end.  If  each  bar  is 
29  ft.  long,  what  is  the  total  length  ? 

3.  A  steam  fitter  found  that  the  weight  per  foot  of  3{" 
wrought  iron  pipe  is  9  lb. ;  what  is  the  weight  of  a  piece  of 
3k"  wrought  iron  pipe  12  feet  long  ? 

4.  An  accident  happened  in  a  mill.  A  number  of  men  were 
sent  out  to  make  the  repairs.  The  following  number  of  hours 
was  reported : 

8  men  10  hours  each  3  men    5  hours  each 

4  men  65  hours  each  4  men  21  hours  each 

7  men  14  hours  each  6  men  11  hours  each 

What  was  the  total  number  of  hours  worked? 

5.  The  consumption  of  water  for  a  city  during  the  month  of 
December  was  116,891,213  gallons  and  for  January  115,819,729 
gallons.     How  much  was  the  decrease  in  consumption  ? 

6.  An  order  to  a  machine  shop  called  for  598  machines  each 
weighing  1219  pounds.     What  was  the  total  weight  ? 

7.  If  an  I  beam  weighs  24  lb.  per  foot  of  length,  find  the 
weight  of  one  measuring  16'  9"  long. 

8.  Multiply  641  and  225. 

9.  Divide  24,566  by  319. 

10.  An  order  was  sent  for  ties  for  a  railroad  847  miles  long. 
If  each  mile  required  3017  ties,  how  many  ties  would  be 
needed  ? 

11.  How  many  gallons  j)er  minute  are  discharged  by  two 
pipes  if  one  discharges  25  gallons  per  minute  and  the  other 
6  gallons  less  ? 


12  VOCATIONAL  MATHEMATICS 

Factors  ' 

The  factors  of  a  number  are  the  integers  which  when  multi- 
plied together  produce  that  number. 

Thus,  21  is  the  product  of  3  and  7  ;  hence,  3  and  7  are  the  factors  of  21. 
Separating  a  number  into  its  factors  is  (tailed  factoring. 
A  number  that  has  no  factors  but  itself  and  1  is  a  prime 
number. 

The  prime  numbers  up  to  25  are  2,  3,  5,  7,  11,  13,  17,  19  and  23. 
A  prime  number  used  as  a  factor  is  2i  prime  factor. 
Thus,  3  and  5  are  prime  factors  of  15. 

Every  prime  number  except  2  and  5  ends  with  1,  3,  7,  or  9. 
To  find  the  prime  factors  of  a  number. 
Example.  —  Find  the  prime  factors  of  84. 

2)84  The  prime  number  2  divides  84  evenly,  leaving  the  quotient 

2)42  ^'^'  which  2  divides  evenly.  The„next  quotient  is  21  which  3 
divides,  giving  a  quotient  7.  7  divided  by  7  gives  the  last 
quotient  1  which  is  indivisible.  The  several  divisors  are  the 
prime  factors.  So  2,  2,  3,  and  7  are  the  prime  factors 
of  84. 


3)21 

7)7 
1 


Phoof.  —  The  product  of  the  prime  factors  gives  the  number. 
EXAMPLES 

Find  the  prime  factors : 

1.  63                           4.   636  7.  1155 

2.  60                            5.   1572  8.  7007 

3.  250                          6.   2800  9.  13104 

Cancellation 

To  reject  a  factor  from  a  number  divides  the  number  by  that 
factor ;  to  reject  the  same  factors  from  both  dividend  and  divisor 
does  not  affect  the  quotient.    This  process  is  called  cancellation. 

This  method  can  be  used  to  advantage  in  many  everyday  cal- 
culations. 

Example,  —  Divide  12  x  18  x  30  by  6  x  9  x  4. 


REVIEW  OF  ARITHMETIC  13 

1 

2        ;2      16  By  ^^^  method  it  is  not 

Dividend  ;S2  X  ;?  X  3P      OA  ^     ..     .       neccBsary  to  multiply  be- 

Tk'   •  a  w Q  w  7    =  ^^  Quotient.      fore    dividing.       Indicate 

\^      ^      '  the    division    by    writing 

r  the  divisor  under  the  divi- 

1  dend  with  a  line  between. 

Since  6  is  a  factor  of  6 

and  12,  and  9  of  9  and  18,  reapectively,  they  may  be  cancelled  from  both 

divisor  and  dividend.     Since  2  in  the  dividend  is  a  factor  of  4  in  the 

divisor  it  may  be  cancelled  from  both,  leaving  2  in  the  divisor.     Then  the 

2  being  a  factor  of  30  in  the  dividend,  is  cancelled  from  both,  leaving  15. 

The  product  of  the  uncancelled  factors  is  30.     Therefore,  the  quotient 

is  30. 

Proof.  —  If  the  product  of  the  divisor  and  the  quotient  equal  the 
dividend,  the  answer  is  correct. 

EXAMPLES 
Indicate  and  find  quotients  by  cancellation : 

1.  Divide  36  X  27  x  49  x  38  x  50  by  70  x  18  x  15. 

2.  What  is  the  quotient  of  36  x  48  X  16  divided  by  27  x  24 
X8? 

3.  How  many  pounds  of  washers  at  50  cents  a  pound  must 
be  given  in  exchange  for  15  pounds  of  bolts  at  40  cents  a 
pound  ? 

4.  There  are  16  ounces  in  a  pound  ;  30  pounds  of  steel  will 
produce  how  many  horseshoes,  if  each  weighs  6  ounces  ? 

5.  How  many  pieces  of  steel  rod,  each  weighing  10  pounds, 
at  20  cents  a  pound,  must  be  given  in  exchange  for  10  bars  of 
jY'  iron  rod,  each  weighing  5  pounds,  at  4  cents  per  pound  ? 

6.  Divide  the  product  of  10,  75,  9,  and  96  by  the  product  of 
5,  12,  15,  and  9. 

7.  If  24  men,  working  9  hours  a  day,  can  do  a  piece  of  work 
in  12  days,  how  many  days  will  it  take  18  men,  working  8 
hours  a  day,  to  do  the  work  ? 


14  VOCATIONAL  MATHEMATICS 


Greatest  Common  Divisor 

The  greatest  common  divisor  of  two  or  more  numbers  is  the 
greatest  number  that  will  exactly  divide  each  of  the  numbers. 

To  find  the  greatest  common  divisor  of  two  or  more  numbers. 

Example.  —  Find  the  greatest  common  divisor  of  90  and 
150. 

90  =  2x3x5x3          2)90  150                 First  Method 

150  =  2  X3x5x5            5)45  75           The  prime  factors  com- 

Ans.  30  =  2  X  3  X  5                    3)9  15      ^^^^  ^o  both  90  and  150 

"o  K       are  2,  3,  and  5.     Since 

2  X  3  X  5  =  30  Ans.  ^^^  greatest  common  di- 

visor  of  two  or  more  num- 

90)150(1  hers   is   the   product   of 

gQ  their  common  factors,  30 

TTT^NQp..-.  is   the   greatest  common 

' ^^^  divisor  of  90  and  150. 

bO 

Greatest  Common  Divisor  30)60(2  '^^^^^^  ^^^^^^ 

(jQ  To    find    the    greatest 

—  common     divisor     when 

the  numbers  cannot  jbe  readily  factored,  divide  the  larger  by  the  smaller, 
then  the  last  divisor  by  the  last  remainder  until  there  is  no  remainder. 
The  last  divisor  will  be  the  greatest  common  divisor.  If  the  greatest  com- 
mon divisor  is  to  be  found  of  more  than  two  numbers,  find  the  greatest 
common  divisor  of  two  of  them,  then  of  this  divisor  and  the  third  num- 
ber, and.  so  on.  The  last  divisor  will  be  the  greatest  common  divisor  of 
all  of  them. 

EXAMPLES 
Find  the  greatest  common  divisor : 

1.  270,810.  3.   504,560.  5.   72,153,315,2187. 

2.  264,  312.  4.   288,  432,  1152. 


Least  Common  Multiple 

The  product  of  two  or  more  numbers  is  called  a  multiple  of 
each  of  them;   4,  6,  8,  12  are  multiples  of  2.     The  multiple 


RKVIKW   OF   AHITHMKTIC  15 

of  two  or  more  numbers  is  called  the  common  multiple  of  the 
numbers ;  00  is  a  common  multiple  of  4,  5,  6. 

The  least  common  multiple  of  two  or  more  numbers  is  the 
smallest  common  multiple  of  the  number;  30  is  the  least 
common  multiple  of  8,  5,  6. 

To  find  the  least  common  multiple  of  two  or  more  numbers. 

ExAMPLK.  —  Find   the   least   common   multiple   of  21,   28, 

and  30.  _..   ,  ,. ,.   ^ 

,  First  Method 

-1=3x7  Take  all  the  factors  of  the  firet  number,  all  of 

28  =  2  X  2  X  7      ^h^  second  not  already  represented  in  the  first,  etc. 
30  =  2  X  i}  X  5      'f  has, 

3  x7x2x2x5  =  420i.C.3f. 

Second  Method 

2)21     28     30 

3)21     14     15 

7)7     14       5 


2  X  3  x  7  X  1  x  2  X  5  =  420  X.  (7.  M. 

Divide  any  two  or  more  numbers  by  a  prime  factor  contained  in  them, 
like  2  in  28  and  30.  Write  21  which  is  not  divided  by  the  2  for  the  next 
quotient  together  with  tl>e  14  and  16.  3  is  a  prime  factor  of  21  and  15 
which  gives  a  quotient  of  7  and  5  with  14  written  in  the  quotient  undi- 
vided. 7  is  a  prime  factor  of  7  and  14  which  gives  a  remainder  of  1,  2  ; 
and  6  undivided  is  written  down  as  before.  The  product  420  of  all  these 
divisors  and  the  last  quotients  i^  the  least  common  multiple  of  21,  28, 
and  30. 

EXAMPLES 

Find  the  least  common  multiple : 

1.   18,  27,  30.        2.   15,  60,  140,  210.  3.   24,  42,  54,  360. 

4.   25,  20,  35,  40.  5.   24,  48,  96,  192. 

6.  What  is  the  shortest  length  of  rope  that  can  be  cut  into 
pieces  32',  36',  and  44'  long? 


16  VOCATIONAL  MATHEMATICS 

Fractions 

A  fraction  is  one  or  more  equal  parts  of  a  unit.  If  an  apple 
be  divided  into  two  equal  parts,  each  part  is  one-half  of  the 
apple,  and  is  expressed  by  placing  the  number  1  above  the 
number  2  with  a  short  line  between :  i  A  fraction  always 
indicates  division.  In  i,  1  is  the  dividend  and  2  the  divisor ; 
1  is  called  the  nuinerator  and  2  is  called  the  denominator. 

A  common  fraction  is  one  which  is  expressed  by  a  numerator 
written  above  a  line  and  a  denominator  below.  The  nu- 
merator and  denominator  are  called  the  terms  of  the  fraction. 

A  proper  fraction  is  a  fraction  whose  value  is  less  than  1 ;  its 
numerator  is  less  than  its  denominator,  as  -|,  -f,  f,  ^.  An 
improper  fraction  is  a  fraction  whose  value  is  1  or  more  than  1; 
its  numerator  is  equal  to  or  greater  than  its  denominator,  as  f, 
■}-f.  A  number  made  up  of  an  integer  and  a  fraction  is  a 
mixed  number.  Read  with  the  word  and  between  the  whole 
number  and  the  fraction :  4:^\,  S^,  etc. 

The  value  of  a  fraction  is  the  quotient  of  the  numerator 
divided  by  the  denominator. 

EXERCISE 
Read  the  following : 

1.   I  3.    121-  5.   ^  7.   9^  9.   I 

2.  \i     4.  ^       6.  ^      8. 12^ 

Reduction  of  Fractions 

Reduction  of  fractions  is  the  process  of  changing  their  form 
without  changing  their  value. 

To  reduce  a  fraction  to  higher  terms. 

Multiplying  the  denominator  and  the  numerator  of  the  given 
fraction  by  the  same  number  does  not  change  the  value  of  the 
fraction. 


REVIEW   OF  ARITHMETIC  17 

Example.  —  Reduce  |  to  thirty -seconds. 

The  denominator  must  be  multiplied  by  4  to 
5  ^1  _  ^  jins,  obtain  32  ;  so  the  numerator  must  be  multiplied 
8     4       32  by  the  same  number  so  that  the  value  of  the 

fraction  may  not  be  changed. 

EXAMPLES 
Change  the  following : 

1.  f  to  27ths.  6.  ^^  to  75th8. 

2.  \^  to  60ths.  7.  Jl  to  144ths. 

3.  f  to  40ths.  8.  ^  to  168ths.  - 

4.  Jto56ths.  9.  |Jto522ds. 

5.  3%  to  50ths.  10.  ^  to  9375ths. 

A  fraction  is  said  to  be  in  its  lowest  terms  when  the  numersr 
tor  and  the  denominator  are  prime  to  each  other. 

To  reduce  a  fraction  to  its  lowest  terms. 

Dividing  the  numerator  and  the  denominator  of  a  fraction 
by  the  same  number  does  not  change  the  value  of  the  fraction. 
The  process  of  dividing  the  numerator  and  denominator  of  a 
fraction  by  a  number  common  to  both  may  be  continued  until 
the  terms  are  prime  to  each  other.' 

Example.  —  Reduce  -f|  to  fourths. 

The  denominator  must  be  divided  by  4  to  give 
12  _  3    J,  the  new  denominator  4  ;  then  the  numerator  must  be 

16      4  divided  by  the  same  number  so  as  not  to  change  the 

value  of  the  fraction. 

If  the  terms  of  a  fraction  are  large  numbers,  find  their 
greatest  common  divisor  and  divide  both  terms  by  that. 

Example.  —  Reduce  |Jff  to  fourths. 

(1)  2166)2888(1  (2)  2166^3     . 

2166  2888     4      ^  ' 

O.  a  D.       722)2166(3 
2166 


18  VOCATIONAL   MATHEMATICS 

EXAMPLES 
Reduce  to  lowest  terms  : 

1-  A         3-  Hil        s-  H         '•  IS*         9-  Hi 

2-    m  *■    U  6.    -rVj  8.    iff  10.    T-VA 

To  reduce  an  integer  to  an  improper  fraction. 
Example.  —  Reduce  25  to  fifths. 

25times4  =  l*i   Ans.      ,   '"  '  "'f""'[!,^-    ^  25  there  must  be 
5  5  25  tunes  5,  or  i|^. 

To  reduce  a  mixed  number  to  an  improper  fraction. 

Example.  —  Reduce  16^  to  an  improper  fraction. 

i_  sevenths  Since  in  1  there  are  },  in  16  there  must 

112  be  16  times  ^,  or  ^2^ 

4  sevenths  lp  +  f=^K 

116  sevenths,  =  J-^. 

EXAMPLES 
Reduce  to  improper  fractions  : 

1.  3J  3.   171  5.   13J  7.   S59j% 

2.  16^  4.   12^  6.   27t«^  8.   482i| 
9.   25-^                                 -  10.   Reduce  250  to  16ths. 

11.  Change  156  to  a  fraction  whose  denominator  shall  be  12. 

12.  In  $  730  how  many  fourths  of  a  dollar  ? 

13.  Change  12 1  to  16ths.  14.    Change  24f  to  ISths. 

To  reduce  an  improj^er  fraction  to  an  integer  or  mixed  number 
divide  the  numerator  by  the  denominator. 

Example.  —  Reduce  -\%^-  to  an  integer  or  mixed  number. 
24 


16)385 

^^  Since  ||  equal  1,  ^^  will  equal  as  many 

times  1  as  16  is  contained  in  385,  or  24^15 

65         24-jL   Ans.      ^^^^^^^ 

64 

1 


REVIEW  OF  ARITHMETIC  19 

EXAMPLES 
Reduce  to  intecjers  oi*  mixed  numbers : 

1.  H  *•  'iV  7.  VV  10.  i^i 

.  Wlieii  fractions  have  the  same  denominator  their  denomi- 
nator is  called  a  common  denominator. 
Thus,  |§,  ^,  ^j  have  a  common  denominator. 

The  smallest  common  denominator  of  two  or  more  fractions  is 
their  least  common  denominator. 

Thus,  \^,  ^«2,  ^  become  ^,  |,  J  when  changed  to  their  least  common 
denominator. 

The  common  denominator  of  two  or  more  fractions  is  a 
common  multiple  of  their  denominators. 

The  least  common  denominator  of  two  or  more  fractions  is 
the  least  common  multiple  of  their  denominators. 

Example.  — Reduce  }  and  f  to  fractipi^s  having  a  common 

denominator. 

a  w  6  _-  18  The  common    denominator  must  be  a 

,       4  _  20  common  multiple  of  the  denominators  4 

^       T  ~  tT  and  0,  and  since  24  i.s  the  product  of  the 

?  ~  Ti  ^"^  t  —  TX  denominators,  it  is  a  common  multiple  of 

them.     Therefore.  24  is  a  common  denominator  of  J  and  f . 

To  reduce  fractions  to  fractions  having  the  least  common  denominator. 

Example.  —  Reduce  |,  |,  and  j^  to  fractions  having  the 
least  common  denominator. 

2^^  S     6     12  "^'^^   least  common   de- 

^— — —  nominator    must    be    the 

1 — z.  least  common  multiple  of 

1      1        -^  the  denominators  3,  6,  12, 

2x3x2  =  12    L.  r'.  M.  which  is  12. 

f  =  -(8^ ;    ^  =  1^  ;    Vj  =  f^.      Ans.  l^'vide  the  least  common 

multiple  12  by  the  denom- 
inator of  each  fraction,  and  multiply  both  terms  by  the  quotient.     If  the 


20  VOCATIONAL  MATHEMATICS 

denominators  should  be  prime  to  each  other,  their  product  would  be  theii 
least  common  denominator. 


EXAMPLES 
Reduce  to  fractions  having  a  common  denominator  : 

1-  h  i  5-  f>  I,  I 

4-    f ,  tV,  i  8-    i>  A,  I,  i 

Reduce  to  fractions  having  least  common  denominator : 

1-  I,  J,  A  5-    1,1,  A,  4 

2-  I,  h  A  6.    I,  f ,  I,  I 

^-   A>  TO  i  7.   Which  fraction  is  larger, 

4-   iA.A.A  fori? 

Addition  of  Fractions 

Only  fractions  with  a  common  denominator  can  be  added. 
If  the  fractions  have  not  the  same  denominator,  reduce  them 
to  a  common  denominator,  add  their  numerators,  and  place 
their  sum  over  the  common  denominator.  The  result  should  be 
reduced  to  its  lowest  terms.  If  the  result  is  an  improper 
fraction,  it  should  be  reduced  to  an  integer  or  mixed  number. 


The        least 
common  multi- 


ExAMPLE.  —  Add  I,  |,  and  y\. 
1.   2)4     6     16 

2}2__3__8  pie  of  the  de- 

13       4     48  Z/.  C.  M.  nominators     is 

o     3i5i9    36i40i27_i03        J*n,       48.        Dividing 

this  by  the  de- 
nominator of  each  fraction  and  multiplying  both  terms  by  the  quotient 
give  If,  f§,  f|.  The  fractions  are  now  like  fractions,  and  are  added  by 
adding  their  numerators  and  placing  the  sum  over  the  common  denomi- 
nator.    Hence,  the  sum  is  ^5^/-,  or  2j7j. 


REVIEW  OF  ARITHMETIC  21 

Example.  —  Add  5J,  7y^,  and  6^j^- 

^i    =    ^a  First  find  the  sum  of  the  fractions, 

7^=    TfJ-  which  is  fj,  or  1|^.     Add  this  to  the 

QJL  =    6lr4t  sum  of  the  integers,  18.     18  -|-  1 J^  = 

i8|f=19H.  Ana.      ^m- 

EXAMPLES 

1.  Find  the  "over-all"  dimension  of  a  drawing  if  the 
separate  parts  measure  -j^",  f",  ^",  and  -^"j  respectively. 

2.  Find  the  sum  of  ^,  J,  J,  j-f ,  and  fj. 
a   Find  the  sura  of  3|,  4},  and  2^. 

4.  Four  castings  weigh  respectively  8J  lb.,  5^  lb.,  llf  lb., 
and  7|  lb.     What  is  their  total  weight  ? 

5.  The  diameter  of  two  holes  is  3|"  and  the  distance  be- 
tween the  sides  of  the  holes  is  J".  What  is  the  distance  from 
the  outside  of  one  hole  to  the  outside  of  the  other  ? 

6.  Two  brass  rods  measure  8-j^"  and  5^"  What  is  their 
combined  length  ? 

7.  A  board  was  cut  into  two  pieces,  one  8f"  and  the  other 
6^"  long.  If  ^"  was  allowed  for  waste  in  cutting,  what  was 
the  length  of  the  board  ? 

8.  Three  pieces  of  rod  contain  38^,  12^,  and  53|  feet  re- 
spectively.    What  is  their  total  length  in  feet? 

9.  Add:  lOi,  7|,  11,  i|. 

Subtraction  of  Fractions 

Only  fractions  with  a  common  denominator  can  be  sub- 
tracted. If  the  fractions  have  not  the  same  denominator, 
reduce  them  to  a  common  denominator,  and  write  the  differ- 
ence of  their  numerators  over  the  common  denominator.  The 
result  should  be  reduced  to  its  lowest  terms. 


22  VOCATIONAL  MATHEMATICS 

Example.  —  Subtract  J  from  |. 

5_2_i5._i2_    3  The  least  common  denominator  of 

^    _/  =  i'    ^]^.      ^  I    and    I    is    6.      t  =  f,    and    f  =  ^. 

^^       ^'  '  Their  difference  is  \. 

Example.  —  From  11^  subtract  5|. 

11 1  ^  10  8  When  the  fractions  are  changed  to 

^ 5^  __    ^5  their  least  common  denominator,  they 

^       — ^  _  fii        A     '  ^^^  ^^^  ~  ^t-     I  cannot  be  subtracted 

*>  ~     2*  •  •         from  I,  hence  1  is  taken  from  11  units, 

changed  to  sixths,  and  added  to  the  f  which  makes  f .     10|  —  4|  =  6|=6|. 


EXAMPLES 

1.  The  distance  between  two  holes  is  5f "  measured  from  the 
centers.  If  the  holes  are  -^^"  in  diameter,  what  is  the  length 
of  metal  plate  between  them  ? 

2.  From  a  steel  bar  26|"  long  were  cut  the  following  pieces : 
one  7\",  one  6^",  one  3|"  long.  If  after  cutting  these  pieces, 
the  length  of  the  bar  was  8f",  what  was  the  amount  of  waste 
in  cutting? 

3.  A  piece  of  steel  on  a  lathe  is  1"  in  diameter.  In  the  first 
cut  -^"  were  taken  off,  in  the  second  cut  g-\",  in  the  third  cut 
-jig-",  in  the  fourth  -^-q".  What  was  the  diameter  of  the  finished 
piece  ? 

4.  A  bolt  is  15^"  long.  How  much  must  be  cut  from  it  to 
make  it  llf "  long  ? 

5.  In  wiring  a  house  five  men  work  14  hours ;  one  man  works 
1  hour  and  20  minutes ;  a  second  man  works  2  hours  and  15 
minutes ;  a  third  man  works  5  hours ;  and  a  fourth  man,  4J 
hours.     How  many  hours  did  the  fifth  man  work  ? 

6.  It  is  S\"  between  the  centers  of  two  holes  of  the  same 
size.  The  distance  between  the  sides  of  the  holes  is  1^" 
What  k  the  diameter  gf  each  hole  ? 


REVIEW  OF  ARITHMETIC  23 

7.  There  were  48^  gallons  in  the  tank.  First  4^  gallons 
were  used,  then  5^  gallons,  and  last  2'j  gallons.  How  many 
gallons  were  left  in  the  tank  ? 

8.  \Vhat  is  the  difference  between  ^  and  JJ  ? 

9.  What  is  the  difference  between  32J  and  .SJJ  ? 

10.  An  electrician  had  a  reel  of  .SOO  feet  of  copper  wire.  He 
used  at  various  times  50^',  32 J',  1091',  and  2737^'.  How  much 
wire  was  left? 

Multiplication  of  Fractions 

To  multiply  fnictiona,  mnltiply  the  numerators  together  for  the 
neiv  numerator  and  multiply  the  denominators  together  /or  the 
new  denominator. 

Cancel  when  possible.  The  word  of  between  two  fractions 
is  equivalent  to  the  sign  of  multiplication. 

To  multiply  a  mixed  number  by  an  integer,  nmltiply  the  whole 
number  and  the  fraction  separately  by  the  integer  then  add  the 
prrxlucts. 

To  multiply  two  mixed  numbers,  change  each  to  an  improper 
fraction  and  multiply. 

Example.  —  Multiply  |  by  J. 

I  multiplied  by  J  is  the  same  as  5  0/ 1.  3  and  5  are  prime  to  each  other 
so  that  answer  is  |.  This  method  of  solution  is  the  same  as  multiplying 
the  numerators  together  for  a  new  numerator  and  the  denominators  for 
a  new  denominator.     Cancellation  shortens  the  process. 

Example.  —  Find  the  product  of  124}  and  5. 

124} 

-  If  the  fi-action  and  integer  are  mul- 

— ^  «;      4  —  L5  —  <ii       tiplied  separately  by  5,  the  result  is  6 

3}  oxj-'^-J       ^.^^^^  J  =  V  =  ^^  and  6  times  124  = 

620  620.     620  +  3f  =G23}. 

623}  Ana. 


24  VOCATIONAL  MATHEMATICS 

EXAMPLES 

1.  What  is  the  length  of  a  bar  of  iron  that  has  been  cut  into 
8  pieces,  each  j\"  in  length  ? 

2.  What  is  ^  of  J  ? 

3.  What  is  the  cost  of  5^  lb.  of  iron  castings  at  4J  cents  a 
pound  ? 

4.  If  a  car  is  filled  with  machinery  of  a  certain  kind  in  4^ 
hours,  how  long  will  it  take  to  fill  10  cars  ? 

5.  If  I  of  the  shell  of  a  stationary  boiler  is  considered  as  the 
heating  surface,  how  many  square  feet  of  heating  surface  are 
there  in  a  boiler  containing  9S^  sq.  ft.  ? 

6.  Multiply  5/^  by  3^. 

7.  What  is  the  cost  of  4|  pounds  of  castings  for  making 
bench  lathes  at  3^  cents  per  pound  ? 

8.  What  is  the  cost  of  5^  lb.  of  hammered  copper  at  31|- 
cents  per  pound  ? 

9.  If  j  lb.  of  bolts  are  used  to  make  a  small  machine,  how 
many  pounds  of  bolts  must  be  made  for  14  such  machines  ? 

10.  If  10|  bundles  of  shingles  are  used  on  one  side  of  a 
square  hip-roof,  how  many  bundles  will  the  whole  roof  require  ? 

Division  of  Fractions 

To  divide  one  fraction  by  another,  invert  the  divisor  and 
proceed  as  in  multiplication  of  fractions.  Change  integers  and 
mixed  numbers  to  improper  fractions. 

Example.  —  Divide  |  X  f  by  f  x  f . 
*xf  +  (fx|)  = 

2  The  divisor  f  x  f  is  inverted  and  the 

^      3      0      ^  _  ^     /J  result  obtained  by  the  process  of  cancel- 

5^^''^'^^~5  lation. 


REVIEW  OF  ARITHMETIC  25 

Example.  —  Divide  3156|  by  5. 

6)3166|  y^^^^  ^j^^  integer  of  a  mixed 


30 


number    is    large,    it    may    be 


15  divided  as  follows:  5  in  3166}, 

15  1|  =  J                              631  times,  with  a  remainder  of 

^  If.     This  remainder  divided  by 

g  5  gives  ,'<),  which. is  placed  at 


the  right  of  the  quotient. 


Example.  — Divide  3682  by  5}. 

When  the  dividend  is  a  large  number  and 
5V)  3682  the  divisor  a  mixed  number  as  in  3682  -j-  6J, 

9  2  multiply  both  dividend  and  divisor  by  2,  when 

T?  \-Q/>  i  ^^^  divisor  becomes  11  halves  and  the  dividend 

^ '        7364  halves.     Multiplying  both  dividend  and 

""^TT    Ans.      divisor  by  the  same  number  does  not  change 
the  quotient.     Dividing,  the  quotient  is  6603^. 

A  fraction  having  a  fraction  for  one  or  both  of  its  terras  is 
called  a  complex  fraction. 

To  reduce  a  complex  fraction  to  a  simple  fraction. 

42 
Example.  — Reduce  ^  to  a  simple  fraction. 

||  =  ^  =  ¥^V-  =  ¥x^=if  Ans. 

Change  4f  and  7^  to  improper  fractions,  ^*  and  V,  respectively.  Per- 
form the  division  indicated  with  the  aid  of  cancellation  and  the  result  will 
bejf 

EXAMPLES 


1.   Divide  ^  by  |. 

7.   296-10i  =  ? 

2.   Divide  ^^  by  f . 

8.   28,769 -^7f=? 

3.    Divide  |f  by  \. 

•■?r' 

4.   Divide  J  by  \. 

5.  Divide  }  by  J. 

6.  384J^5  =  ? 

-  -m-' 

26 


VOCATIONAL  MATHEMATICS 


Drill  in  the  Use  of  Fractions 


Addition 


1. 

i  +  4  =? 

19. 

1 

8 

+  i     =?     . 

37. 

A 

+  i  =? 

2. 

i  +  \  =^ 

20. 

i 

+  1     =? 

38. 

A 

+  i  =? 

3. 

i  +  i  ='' 

21. 

i 

+  i     =? 

39. 

iV 

+  i  =? 

4. 

i  +  TV  =  ? 

22. 

i 

+  ,v  =  ? 

40. 

A 

+  tV  =  ? 

5. 

i+^=? 

23. 

I 

+  .v=? 

41. 

3V 

+  A=? 

6. 

i+^v=? 

24. 

1 

+  6^4=? 

42. 

A 

+  6V  =  ? 

7. 

i+i  =? 

25. 

tV 

+  i     =? 

43. 

A 

+4  =? 

8. 

i  +  i  =^ 

26 

tV 

+  i     =? 

44. 

eV 

+i  =? 

9. 

i  +  i  =? 

27. 

tV 

+  i     =? 

4S. 

A 

+i  =? 

10. 

i  +  iV  =  ? 

28. 

tV 

+  tV  =  ''' 

46. 

A 

+  tV  =  ? 

11. 

i  +  A  =  ^ 

29. 

-A 

+  A=? 

47. 

A 

+A=? 

12. 

i+6V  =  '^ 

30. 

t's 

+  «V=? 

48. 

A 

+A=? 

13. 

l  +  i  =? 

31. 

i 

+  4  =? 

49. 

* 

+1  =? 

14. 

l  +  i  =? 

32. 

3 

+  i  =? 

SO. 

i 

+  i  =? 

15. 

l  +  i  =? 

33. 

i 

+  i  =? 

51. 

i 

+  i  =? 

16. 

I  +  tV  =  ? 

34. 

i 

+  tV  =  ? 

52. 

1 

+  tV  =  ? 

17. 

I  +  ^V  =  ? 

35. 

8 
4 

+  A  =  ? 

53. 

i 

+A=? 

18. 

l  +  -6^=? 

36. 

f 

+  eV  =  ? 

54. 

i 

+A=? 

Subtraction 

1. 

i-i  =? 

8. 

i- 

-i  =? 

15. 

5 

8" 

-i  =? 

2. 

i-i  =? 

9. 

i- 

-i  =? 

16. 

i 

-tV  =  ? 

3. 

i-i  =? 

10. 

\ 

-tV  =  ? 

17. 

1- 

-A=? 

4. 

i-TV=? 

11. 

\- 

-sV  =  ? 

18. 

f- 

-Vt  =  ? 

5. 

i-^V  =  ? 

12. 

i- 

-^v=? 

19. 

i- 

-f  =? 

6. 

i-^V  =  ? 

13. 

-4  =-? 

20. 

i- 

-A  =  ? 

7. 

i-^=? 

14. 

f 

-i  =? 

21. 

1 ' 

-i  =? 

REVIEW  OF  ARITHMETIC 


27 


22-  i  -.',  =  ? 

23.  i  -3'l  =  '' 

24.  i  -ff'4  =  -" 
25-4  -T%=-' 

26.  ^  -tV=- 

27.  A\--tV  =  '^ 

28.  -jV  —  tV  =  ^' 
29-  tV  -  /2  =  ■•• 

30.  T*! 

31.  f 

32.  f 


1 

ITT 


1      —  .» 


-i     = 


1.  4 

2.  ^ 

3-  4 

4.  * 


6.  i 

a  i 

9.  i 

10.  J 

11.  J 

12.  i 

13.  I 

14.  f 

15.  I 

16.  J 

17.  » 

18.  I 


X  J     —  . 

xi    =? 

X^  =? 
XA  =  ? 

V  1      *> 

xi  =? 
xi  =? 
xi  =? 
Xt',  =  '^ 

x,V  =  ^^ 

X^  =  ? 
Xi  =? 
X  J  =? 
xj   .=  ? 

xA  =  ? 

V  1.  —  9 


33.  J  -J     = 

34.  3  -l'«  =  -' 

35.  J  -5>j  =  V 

36.  I  -j^f  = 

37.  U-jV  = 

38.  i    -A  = 

39.  H-A  = 

40.  H-A  = 

42.  i?-s^  = 

43.  4    -«  = 

Multiplication 
19.    I    X  .V    =  ? 

__  9 


20.    J     xi    =? 


21.  i     xi 

22.  J    X  j5 


23. 


i     X^j  =  ? 

1  s^        \       •> 


24.    i     Xj't 


44.  1     -A  = 

45.  \    -i,= 

46.  t*-."!" 

47.  s't-j^  = 

48.  A-,'i  = 

49.  i     -i    = 

50.  J    -V    = 

51-  i    -i     = 

52.  J     -t'j  = 

53.  J    -A  = 

54.  i    -A  =  ? 


37.  ,V 

38.  ,>j 

39.  ^j 

40.  ,V 

41.  ^j 

42.  ^2 


25.  ^Vxi    =?  43.    ,V 

26.  T»jXi    =?  44.    ,>j 

27.  iVxi    =? 

28.  AXt>j  =  ? 

29.  T'ffX^j  =  ? 

30.  T'irX5'i  =  ? 

31.  I     X  4    =  ? 

32.  J     X  }    =  ? 

33.  I     Xj    =? 

34.  J     X  ,V  =  ? 

35.  f     X  jV  =  '.' 

36.  J     Xji  =  ? 


45.  sV 

46.  ,V 

47.  i^T 

48.  sV 

49.  I 

50.  i 

51.  J 

52.  J 

53.  i 

54.  J 


X4  =? 
X  i  =? 
Xi  =? 
Xt'5  =  ? 
Xi^l  =  V 
X  ^I  =  V 
xi  =? 
X}    =? 

xi    =? 

xA  =  "'' 

x^=? 

X  «V  =  ■'' 
xi  =? 
xi  =? 
xi    =? 

Xt',=  V 

X  j'l  =  '! 
XVj=? 


28 


VOCATIONAL  MATHEMATICS 


Division 


37. 
38. 
39. 
40. 
41. 
42. 
43. 
44. 
45. 
46. 
47. 
48. 
49. 
50. 
51. 
52. 
53. 
54. 


Decimal  Fractions 

A  power  is  the  product  of  equal  factors,  as  10  x  10  =  100. 
10  X  10  X  10  =  1000.  100  is  the  second  power  of  10.  1000  is 
the  third  power  of  10. 

A  decimal  fraction  or  decimal  is  a  fraction  wliose  denominator 
is  10  or  a  power  of  10.  A  common  fraction  may  have  any 
number  for  its  denominator,  but  a  decimal  fraction  must  always 
have  for  its  denominator  10,  or  a  power  of  10.  A  decimal  is 
written  at  the  right  of  a  period  (.),  called  the  decimal  point. 
A  figure  at  the  right  of  a  decimal  point  is  called  a  decimal 
figure. 


1. 

i^i    =? 

19. 

\ 

-^i  =? 

2. 

i^i    =? 

20. 

i 

-i  =? 

3. 

i^i    =? 

21. 

i 

^i  =? 

4. 

i-^TV  =  ? 

22. 

i 

^t'^  =  ? 

5. 

i-^A  =  ? 

23. 

i 

-A  =  ? 

6. 

i^^V=? 

24. 

i 

^7V  =  ? 

7. 

i^i     =? 

25. 

iV 

^i  =? 

8. 

i^i    =? 

26. 

tV 

^i  =? 

9. 

i^i    =? 

27. 

tV 

^i  =? 

10. 

i^A  =  ? 

28. 

tV 

-tV=? 

11. 

i^A  =  ? 

29. 

tV 

^A  =  ? 

12. 

i^A  =  ? 

30. 

1^6 

^F't  =  ? 

13. 

I^i  =? 

31. 

8 

^i  =? 

14. 

I^i  =? 

32. 

1 

-i  =? 

15. 

I^i  =? 

33. 

f 

^i  =? 

16. 

I^tV  =  ? 

34. 

* 

^tV  =  ? 

17. 

I^A  =  ? 

35. 

f 

^A=? 

18. 

I^^  =  ? 

36. 

1 

^TrV  =  ? 

A- 

-i  =? 

^v- 

-i  =? 

A- 

-i  =? 

Ti- 

-tV  =  ? 

W- 

-A  =  ? 

7V- 

-A  =  ? 

^v- 

-i  =? 

.v^ 

-i  =? 

A- 

-i  =? 

A- 

-tV  =  ? 

A- 

-^\  =  -f 

e'l- 

-A  =  ? 

J  - 

-i  =? 

i  - 

-i  =? 

i  - 

-i  =? 

i  - 

-tV  =  ? 

i  - 

-A  =  ? 

i  - 

-A  =  ? 

REVIEW  OF  ARITHMETIC 


29 


A  mixed  decimal  is  an  integer  and  a  decimal ;  as,  16.04. 

To  read  a  decimal,  read  the  decimal  as  an  integer,  and  give 
it  the  denomination  of  the  right-hand  figure.  To  tvrite  a  deci- 
mal, write  the  numerator,  prefixing  ciphers  when  necessary  to 
express  the  denominator,  and  place  the  point  at  the  left. 
There  must  be  as  many  decimal  places  in  the  decimal  as  there 
are  ciphers  in  the  denominator. 


EXAMPLES 
Read  the  following  numbers: 


1. 

.7 

7. 

.4375 

13. 

.0000054 

19. 

9.999999 

2. 

.07 

8. 

.03125 

14. 

35.18006 

20. 

.10016 

3. 

.007 

9. 

.21875 

15. 

.0005 

21. 

.000155 

4. 

.700 

10. 

.90625 

16. 

100.000104 

22. 

.26 

5. 

.125 

11. 

.203125 

17. 

y.1032002 

23. 

.1 

6. 

.0625 

12. 

.234375. 

la 

30.3303303 

24. 

.80062 

Express  decimally : 

1.  Four  tenths. 

2.  Three  hundred  twenty-five  thousandths. 

3.  Seventeen  thousand  two  hundred  eleven  hundred-thou- 
sandths. 

4.  Seventeen  hundredths.  6.   Five  hundredths. 

5.  Fifteen  thousandths.  7.   Six  ten-thousandths. 

8.  Eighteen  and  two  hundred  sixteen  hundred-thousandths. 

9.  One  hundred  twelve  hundred-thousandths. 

10.  10  milliouths.  11.    824  ten-thousandths. 

12.  Twenty-nine  hundredths. 

13.  324  and  one  hundred  twenty-six  millionths. 

14.  7846  hundred-millionths. 


30  VOCATIONAL   MATHEMATICS 

1  c    5_6  3  1  2  123         3  2  8  6  5  4 

17.  One  and  one  tenth. 

18.  One  and  one  hundred-thousandth. 

19.  One  thousand  four  and  twenty-nine  hundredths. 

Reduction  of  Decimals 

Ciphers  annexed  to  a  decimal  do  not  change  the  value  of 
the  decimal;  these  ciphers  are  called  decimal  ciphers.  For 
each  cipher  prefixed  to  a  decimal,  the  value  is  diminished  ten- 
fold. The  denominator  of  a  decimal  —  when  expressed  —  is 
always  1  with  as  many  ciphers  as  there  are  decimal  places  in 
the  decimal. 

To  reduce  a  decimal  to  a  common  fraction. 

Write  the  numerator  of  the  decimal  omitting  the  point  for  the 
numerator  of  the  fraction.  For  the  denominator  write  1  with  as 
many  ciphers  annexed  as  there  are  decimal  places  in  the  decimal. 
Then  reduce  to  lowest  terms. 

Example.  —  Reduce  .25  and  .125  to  common  fractions. 

1  Write  25  for  the  numerator  and 

nK 25  'l^  1  A                 1  for  the  denominator  with  two  O's, 

""  100  ~~  ^00  ~~  4  *            which  makes  j^^-^  ;   -^-^^  reduced  to 

4  lowest  terms  is  \. 

1 

^  oe-        125         J2f>        1      A  '^25  is  reduced  to  a  common  frac- 

.125  = =  ^  -    =  -    Ans.      ^.      .    ^^ 

1000    ^p0j^    ^  ^^^^  ^"  ^^^  ^^"^^  ^^y- 

8 

Example.  —  Reduce  .37^  to  a  common  fraction. 

37i  has  for  its  denominator  1 
3  371 

37i^i^l5^_jL  =  ?    Ans      "^^^^^^'"^"^^^^^^^«I^- 
100      100       2        ;0p       8  '  This    is    a    complex    fraction 

4  which   reduced   to  lowest  terms 

isf. 


REVIEW  OF  ARITHMETIC  31 


EXAMPLES 
Reduce  to  cominon  trartions  : 


1.    .01)375 

6.    2.25 

11. 

.16J 

16. 

m 

2.    .15625 

7.    16.144 

12. 

..S3J 

17. 

.66J 

3.    .015625 

a    25.0000100 

13. 

.06J 

18. 

.36i 

4.    .609.375 

9.    1084.0025 

14. 

.140625 

19. 

.83i 

5.    .578125 

10.    .12.V 

15. 

.984375 

20. 

.62^ 

To  reduce  a  cominon  fraction  to  a  decimal. 

Annex  decimal  ciphers  to  the  numerator  and  divide  by  the  de- 
nominator. Point  off  from  the  right  of  the  quotient  as  many 
places  as  there  are  ciphers  annexed.  If  there  are  not  figures 
enough  in  the  quotient,  prefix  cijthers. 

The  division  will  not  always  be  exact,  i.e.  |  =  .142f  or  .142+. 

Example.  — Reduce  |  to  a  decimal. 

.75 
4)3.00 
28 
20 
I  =  .76 

EXAMPLES 

Reduce  to  decimals : 


1. 

^V 

6. 

i 

u. 

A 

16. 

H 

21. 

-^iji 

2. 

TiTT 

7. 

w 

12 

^icT 

17. 

16i 

22. 

25.12i 

3. 

TSU 

8. 

M 

13. 

T2oJi 

18. 

66| 

23. 

33' 

4. 

i 

9. 

A 

14. 

m 

19. 

U 

24. 

^i 

5. 

i 

10 

jh 

15. 

i\ 

20. 

a 
IT 

25. 

Ti^ 

Addition  of  Decimals 

To  add  decimals^  write  them  ,so  that  their  decimal  points  are  in 
a  column.  Add  as  in  integers,  and  place  the  point  in  the  sum 
directly  under  the  points  above  it. 


32  VOCATIONAL  MATHEMATICS 

Example.— Find  the  sum  of  3.87,2.0983,  5.00831,  .029, 
.831. 

3.87 

2  0983  Place  these  numbers,  one  under  the  other,  with 

n  fW'iS'?!  decimal  points  in  a  column,  and  add  as  in  addition 

of  integers.     The  sum  of  these  numbers  should 

'^^^  have  the  decimal  point  in  the  same  column  as  the 

•831  numbers  that  were  added. 


11.83661    Ans. 


EXAMPLES 


Find  the  sum  : 

1.  5.83,  7.016,  15.0081,  and  18.3184. 

2.  12.031,  0.0894,  12.0084,  and  13.984. 

3.  .0765,  .002478,  .004967,  .0007862,  .17896. 

4.  24.36,  1.358,  .004,  and  1632.1. 

5.  .175,  1.75, 17.5,  175.,  1750. 

6.  1.,  .1,  .01,  .001,  100,  10.,  10.1,  100.001. 

7.  Add  5  tenths;  8063  millionths;  25  hundred-thousandths  ; 
48  thousandths;  17  millionths;  95  ten-millionths ;  5,  and  5 
hundred-thousandths ;  17  ten-thousandths. 

8.  Add  24|,  17},  .0058,  7i,  9^^. 

9.  32.58,  28963.1,  287.531,  76398.9341. 
10.   145.,  14.5, 1.45,  .145,  .0145. 

Subtraction  of  Decimals 

To  subtract  decimals,  ivrite  the  smaller  number  under  the 
larger  icith  the  decimal  point  of  the  subtrahend  directly  under  the 
decimal  point  of  the  minuend.  Subtract  as  in  integers,  and  place 
the  point  directly  under  the  points  above. 

Example.  — Subtract  2.17857  from  4.3257. 

Write  the  lesser  number  under  the  greater, 
4.32570  Minuend  y^^^y^  ^^^  decimal    points  under  each  other 

2.17857  Subtrahend       Add  a  0  to  the  minuend,  4.3257,  to  give  it  the 
2.14713  Remainder       same   denominator  as  the  subtrahend.     Then 
subtract  as  in  subtraction  of  integers.     Write 
the  remainder  with  decimal  point  under  the  other  two  points, 


REVIEW  OF  ARITHMETIC  33 

EXAMPLES 

Subtract : 

1.  69.0364-30.8691  =  ?  3.   .0625  -  .03125  =  ? 

2.  48.7209  - 12.0039  =  ?  4.    .0001 1  -  .000011  =  ? 

5.  10  -  .1 -H  .0001  =  ? 

6.  From  one  thousand  take  five  thousandths. 

7.  Take  17  hundred-thousandths  from  1.2. 

8.  From  17.37^  take  14.16f 

9.  Prove  that  ^  and  .500  are  equal. 

10.   Find  the  difference  between  yYA  ^°^  rlfH* 

Multiplication  of  Decimals 

To  multiply  decimals  jnoceed  as  in  integers,  and  give  to  the 
product  as  many  decimal  figures  as  there  are  in  both  multiplier 
and  multiplicand.  When  there  are  not  figures  enough  in  the 
product,  prefix  ciphers. 

Example.  —  Find  the  product  of  6.8  and  .63. 

6.8     Multiplicand 

.63  Multiplier  ^-^  ^^  ^^®  multiplicand  and  .63  the  multiplier. 

"^777  Their  product  is  4,284  with  three  decimal  figures, 

^  the  number  of  decimal  figures  in  the  multiplier 

and  multiplicand. 

4.284  Product 

Example.  —  Find  the  product  of  .05  and  .3. 

.05     Multiplicand  The  product  of  .05  and  .3  is  .016  with  a  cipher 

.3     Multiplier  prefixed  to  make  the  three  decimal  figures  re- 

.015  Product  quired  in  the  product. 

EXAMPLES 
Find  the  products : 

1.  46.25  X. 125  3.   .015  x  .05 

2.  8.0625  X. 1875  4.   25.863  x  4i 


34  VOCATIONAL  MATHEMATICS 

5.  11.11  X  100  8.    .325  X  12i 

6.  .5625  X  6.28125  9.    .001542  x  .0052 

7.  .326  x  2.78  10.   1.001  x  1.01 

To  multiply  btj  10,  100,  1000,  etc.,  remove  the  point  one  place 
to  the  right  for  each  cipher  m  the  multiplier. 

This  can  be  performed  without  writing  the  multiplier. 

Example.— Multiply  1.625  by  100. 

1.625  x  100  =  162.5 

To  multiply  by  200,  remove  the  point  to  the  right  and  multiply 
by  2. 

Example.  —  Multiply  86.44  by  200. 

86.44. 

2 

17,288 

EXAMPLES 
Find  the  product  of : 

1.  1  thousand  by  one  thousandth. 

2.  1  million  by  one  millionth. 

3.  700  thousands  by  7  hundred-thousandths. 

4.  3.894  x  3000  5.   1.892  x  2000. 

Division  of  Decimals 

To  divide  decimals  proceed  as  in  integers,  and  give  to  the  quo- 
tient as  many  decimal  figures  as  the  number  in  the  dividend  ex- 
ceeds those  in  the  divisor. 

Example.  —  Divide  12.685  by  .5. 

The  number  of  decimal   figures  in 

Divisor  .5)12.685  Dividend       the  quotient,  12.685,  exceeds  the  num- 

25.37    Quotient        ber  of  decimal  figures  in  tlie  divisor,  .5, 

by  two.     So  there  tnust  be  two  deQi- 

pial  figures  in  the  quotient. 


REVIEW  OF  ARITHMETIC 


35 


Example.  —  Divide  899.552  by  192. 

When  the  divisor  is  an  integer* 
the  p<iint  in  the  quotient  sliould  lie 
placed  tiin'ctly  over  tlie  i>oint  in 
tlie  dividend,  and  tli<»  diviMfoii  |H»r- 
forme<l  a«  in  Int^gerM.  Tliirt  may 
be  proved  by  multiplying  divinor 
by  quotient,  whicli  would  give  the 
dividend. 


2.081   Quotient 
Dim8or  192)399.552  Dimleud 
384 
1555 
153G 
192 
192 


Example.  —  Divide  28.78884  by  1.25. 

When 
23.031+      Quotient 


Divisor  1.25.)28.78.884  Diridend 

250_ 

378 

375 

388 

375 


134 
125 

9  Remainder 


the  divixor  contains 
decimal  fl^^urea,  move  tlie  i>oint 
in  both  divisor  and  dividend  as 
many  place.s  U)  the  right  as 
tlicre  are  decimal  placcB  in  tlie 
divisor,  which  is  equivalent  to 
multiplyin<{  l)oth  divisor  and 
dividend  by  the  same  number 
and  does  not  change  the  (}Uo. 
tient.  Then  place  the  point  in 
the  quotient  as  if  the  divisor 
were  an  integer.  In  this  ex- 
ample, the   multiplier  of  both 


dividend  and  divisor  is  100. 


EXAMPLES 
Eind  the  quotients : 

1.  .0625 -.125  5.  1000 -I- .001 

2.  315.432 -.132  6.  2.490  +  .  136 

3.  .75 -.0125  7.  28000  +  16.8 

4.  125 -5- 12^ 


a   1.225  +  4.9 
9.   3.1416  +  27 
10.  8.33  +  6 


To  divide  by  10,  100,  1000,  etc.^  remove  tin-  inimt  nue  place  to 
the  left  for  each  cipher  in  the  didaor. 

7V>  divide  by  200,  remove  the  point  ttco  places  to  the  le/t^  and 
divide  by  2. 


36  VOCATIONAL  MATHEMATICS 

EXAMPLES 
Find  the  quotients : 

1.  38.64-10  6.  865.45-5000 

2.  398.42-1000  7.  38.28-400 

3.  1684.32-1000  8.  2.5-500 

4.  1.155-100  9.  .5-10 

5.  386.54-2000  10.  .001-1000 

Parts  of  100  or  1000 

1.  What  part  of  100  is  12^  ?  25  ?  33^  ? 

2.  What  part  of  1000  is  125  ?  250  ?  333^  ? 

3.  How  much  is  i  of  100  ?     Of  1000? 

4.  How  much  is  J  of  100  ?     Of  1000  ? 

5.  What  is  1  of  100  ?     Of  1000? 

Example.  —  How  much  is  25  times  24  ? 
100  times  24  =  2400. 
25  times  24  =  |^  as  much  as  100  times  24  =  600.     Ans. 

Short  Method  in  Multiplication 
To  multiply  by 

25,    multiply  by  100  and  divide  by  4 ; 

33|,  multiply  by  100  and  divide  by  3 ; 

16|,  multiply  by  100  and  divide  by  6 ; 

121  multiply  by  100  and  divide  by  8  ; 

9,      multiply  by  10  and  subtract  the  multiplicand ; 

11,    if  more  than  two  figures,  multiply  by  10  and  add  the 

multiplicand; 
11,  if  two  figures,  place  the  figure  that  is  their  sum  between 
them. 

63  X  11  =  693  74  X  11  =  814 

Note  that  when  the  sum  of  the  two  figures  exceeds  nine,  the  one  in  the 
tens  place  is  carried  to  the  figure  at  the  left. 


REVIEW  OF  ARITHMETIC  37 

EXAMPLES 
Multiply  by  the  short  process : 

1.  81  by  11  =?  10.  68  by  16J  =  ? 

2.  76  by  33i  =  ?  11.  112  by  11  =  ? 

3.  128  by  12J  =  ?  12.  37  by  11  =  ? 

4.  87  by  11  =  ?  13.  4183  by  11  =  ? 

5.  19  by  9  =  ?  14.  364  by  33^  =  ? 

6.  846  by  11  =  ?  15.  8712  by  12J  =  ? 

7.  88  by  11  =  ?  16.  984  by  16J  =  ? 

8.  19  by  11  =  ?  17.  36  by  25  =  ? 

9.  846byl6J  =  ?  la  30by333J=? 


Aliquot  Parts  of  $1.00 

The  aliquot  parts  of  a  nuinber  are  the  numbers  that  are 
exactly  contained  in  it.  The  aliquot  parts  of  100  are  5,  20, 
12J,  16f,  334,  etc. 

The  monetary  unit  of  the  United  States  is  the  dollar,  con- 
taining one  hundred  cents  which  are  written  decimally. 

25    cents  =  $  ^  =  quarter  dollar 

33^  cents  =  $  ^ 

50    cents  =  $  ^  =  half  dollar 


n 

cents  =  $^^ 

«i 

cents  =  $  3^5 

12^ 

cents  =  $  i 

16J 

cents  =  $  i 

10  mills 

5  cents 

10  cents 

10  dimes 

=  1  cent,  ct.    =  $  .01  or  $  0.01 
=  1  "  nickel "  =  $  .05 
=  1  dime,    d.  =  $  .10 
=  1  dollar,  $  =  $1.00 
10  dollars  =  1  eagle,  E.  =  $  10.00 

Example.  —  What  will  69  drills  cost  at  16J  cents  each  ? 
69  drills  will  cost  69  x  16i  cts.,  or  69  x  $i=V  =  f  Hf  =  -^11.60. 


38  VOCATIONAL  MATHEMATICS 

Example.  —  At  25^  a  box,  ho^v  many  boxes  of  nails  can  be 
bought  lor  $  8.00  ? 

8^i  =  8xf  =  32  boxes.     Ans. 

Review  of  Decimals 

1.  For  work  on  a  job  one  man  receives  $  13.75,  a  second 
man  $  12.45,  a  third  man  $  14.21,  and  a  fourth  man  $  21.85. 
What  is  the  total  amount  paid  for  the  work  ? 

2.  A  pipe  has  an  inside  diameter  of  3.067  inches  and  an 
outside  diameter  of  3.428  inches.  What  is  the  thickness  of 
the  metal  of  the  pipe  ? 

3.  At  21  cts.  a  pound,  what  will  be  the  cost  of  108  castings 
each  weighing  29  lb.  ? 

4.  A  man  receives  $  121.50  for  doing  a  piece  of  work.  He 
gives  $12.25  to  one  of  his  helpers,  and  S  10.50  to  another. 
He  also  pays  $75.75  for  material.  How  much  does  he  make 
on  the  job  ? 

5.  An  automobile  runs  at  the  rate  of  9^  miles  an  hour. 
How  long  will  it  take  it  to  go  from  Lowell  to  Boston,  a  dis- 
tance of  26.51  miles  ? 

6.  A  f  square  steel  bar  weighs  1.914  lb.  per  foot.  What 
will  be  the  cost  of  5000  feet  of  |"  steel  bars,  if  it  costs  $  1.75 
per  100  lb.  ? 

7.  Which  is  cheaper,  and  how  much,  to  have  a  13i  cents 
an  hour  man  take  13J  hours  on  a  piece  of  work,  or  hire  a  17^ 
cents  an  hour  man  who  can  do  it  in  9^  hours  ? 

8.  On  Monday  1725.25  lb.  of  coal  are  used,  on  Tuesday 
2134.43  lb.,  on  Wednesday  1651.21  lb.,  on  Thursday  1821.42 
lb.,  on  Friday  1958.82  lb.,  and  on  Saturday  658.32  lb.  How 
many  pounds  of  coal  were  used  during  the  week  ? 

9.  If  in  the  example  above  there  were  10,433.91  lb.  of 
coal  on  hand  at  the  beginning  of  the  week,  how  much  was  left 
at  the  end  of  the  week  ? 


RKVIEW  OF  ARITHMETIC  30 

10.  A  foot  length  of  }"  round  steel  bar  weighs  1.50^5  lb., 
10-foot  length  of  J"  square  steel  bar  weiglis  19.140  lb.,  1-foot 
length  of  y^"  round  steel  bar  weighs  3.017  lb.  What  is  the 
total  weight  of  the  three  pieces  ? 

U.  An  alloy  is  made  of  copper  and  zinc.  If  .(iO  is  copper 
and  .34  is  zinc,  how  many  pounds  of  zinc  and  how  many 
pounds  of  co})per  will  there  be  in  a  casting  of  the  alloy 
weighing  98  lb.  ? 

12..  A  train  leaves  New  York  at  2.10  p.m.,  and  arrives  in 
Philadelphia  at  4.15  p.m.  The  distance  is  90  miles.  What  is 
the  average  rate  per  hour  of  the  train  ? 

13.  The  weight  of  a  foot  of  ^^"  steel  bar  is  1.08  lb.  Find 
the  weight  of  a  21-foot  bar. 

14.  A  steam  pump  pumps  3.38  gallons  of  water  to  each 
stroke  and  the  pump  makes  51.1  strokes  per  minute.  How 
many  gallons  of  water  will  it  pump  in  an  hour? 

15.  At  12^  cents  per  hour,  what  will  be  the  pay  for  23^  days 
if  the  days  are  10  hours  each  ? 

Compound  Numbers 

A  number  which  expresses  only  one  kind  of  concrete  units 
is  a  simple  number;  as,  5  ])k.,  4  knives,  6. 

A  number  composed  of  different  kinds  of  concrete  units  that 
aie  related  to  each  other  is  a  compound  number ;  as,  3  bu.  2  pk. 
1  qt. 

A  denomination  is  a  name  given  to  a  unit  of  measure  or  of 
weight. 

A  number  having  one  or  more  denominations  is  also  called  a 
denominate  number. 

Reduction  is  the  process  of  changing  a  number  from  one 
denomination  to  another  without  changing  its  value. 

Changing  to  a  lower  denomination  is  called  reduction  descend- 
ing; as,  2  bu.  3  pk.  =  88  qt. 


40  VOCATIONAL  MATHEMATICS 

Changing  to   a   higher    denomination   is    called    reduction 
ascending ;  as,  88  qt.  =  2  bu.  3  pk. 

Linear  Measure  is  used  in  measuring  lines  or  distance. 

Table 
12  in.  (in.)  =  1  foot,  ft. 

3  feet  =  1  yard,  yd. 

5^  yards  or  16^  feet  =  1  rod,  rd. 
40  rods  =  1  furlong,  fur. 

8  furlongs  =  1  mile,  mi. 

320  rods,  or  5280  feet  =  1  mile. 
1  mi.  =  320  rd.  =  1760  yd.  =  5280  ft.  =  63,360  in. 

Surveyors'  Measure  is  used  in  measuring  land. 

Table 

7.92  inches  =  1  link,  li. 

100       links    =  1  chain,  ch. 

80      chains  =  1  mile,  mi. 

A  chain,  or  steel  measuring  tape,  100  feet  long,  is  generally  used  by 

engineers.     The  feet  are  usually  divided  into  tenths   instead  of  into 

inches. 

Square  Measure  is  used  in  measuring  surfaces. 

Table 

144    square  inches  =  1  square  foot,  sq.  ft. 

9    square  feet      =  1  square  yard,  sq.  yd. 
30i  square  yards  1       ,  ,  , 

160    square  rods  =  1  acre,  A. 
640    acres  =  1  square  mile,  sq.  mi. 

1  sq.  mi.  =  640  A.  =  102,400  sq.  rd.  =  3,097,600  sq.  yd. 

Cubic  Measure  is  used  in  measuring  volumes  or  solids. 

Table 
1728  cubic  inches  =  1  cubic'foot,  cu.  ft. 

27  cubic  feet  =  1  cubic  yard,  cu.  yd. 

16  cubic  feet  =  1  cord  foot,  cd.  ft. 

8  cord  feet,  or  128  cu.  ft.  =  1  cord,  cd. 
1  cu.  yd.  =  27  cu.  ft.  =  46,666  cu.  in. 


REVIEW  OF  ARITHMETIC  41 

Liquid  Measure  is  used  in  measuring  liquids. 

Table 

4  gills  (gi.)=  1  P'"^  pt" 

2  pints        =  1  quart,  qt. 

4  quart*      =  1  gallon,  gal. 
1  gal.  =  4  qt.  =  8  pt.  =  32  gi. 
A  gallon  contains  231  cubic  inches. 
The  standard  barrel  is  31 J  gal.,  and  the  hogshead  63  gal. 

Dry  Measure  is  used  in  measuring  roots,  grain,  vegetables, 
etc. 

Table 
2  pints    =  1  quart,  qt. 
8  quarts  =  1  peck,  pk. 
4  pecks   =  1  bushel,  bu. 
1  bu.  =  4  pk.  =  32  qt.  =  64  pints. 

The   bushel   contains   2150.42  cubic   inches;    1   dry  quart   contains 
67.2  cu.  in.     A  cubic  foot  is  |J  of  a  bushel. 

Avoirdupois  Weight  is  used  in  weighing  all  common  articles  ; 
as,  coal,  groceries,  hay,  etc. 

Table 

16  ounces  (oz.)        =  1  pound,  lb. 
100  pounds  =  1  hundredweight,  cwt. ; 

or  cental,  ctl. 
20  c\n.,  or  2000  lb.  =  1  ton,  T. 
1  T.  =  20  cwt.  =  2000  lb.  =  32,000  oz. 

The  long  ton  of  2240  pounds  is  used  at  the  United  States  Custom 
House  and  in  weighing  coal  at  the  mines. 

Measure  of  Time. 

Table 
60  seconds  (sec.)  =  1  minute,  min. 
60  minutes  =  1  hour,  hr. 


24  hours 

=  1  day,  da. 

7  days 

=  1  week,  wk. 

365  days 

=  1  year,  yr. 

366  days 

=  1  leap  year. 

100  years 

=  1  century. 

42  VOCATIONAL  MATHEMATICS 

Counting. 

Table 

12  things  =  1  dozen,  doz. 
12  dozen  =  1  gross,  gr. 
12  gross    =  1  great  gross,  G.  gr. 
Paper  Measure. 

Table 
24  sheets  =  1  quire  2  reams     =  1  bundle 

20  quires  =  1  ream  5  bundles  =  1  bale 

Reduction  Descending 
Example.  —  Reduce  17  yd.  2  ft.  9  in.  to  inches. 

1  yd.  =  3  ft. 
17  yd.  =  17  X  3  =  51  ft. 
51  +  2  =  53  ft. 
1  ft.  =  12  in. 
53  ft.  =  53  X  12  =  6.36  in. 
636  +  9  =  645  in.     Ans. 

EXAMPLES 
Reduce  to  lower  denominations : 

1.  46  rd.  4  yd.  2  ft.  to  feet. 

2.  4  A.  15  sq.  rd.  4  sq.  ft.  to  square  inches. 

3.  16  cu.  yd.  25  cu.  ft.  900  cu.  in.  to  cubic  inches. 

4.  15  gal.  3  qt.  1  pt.  to  pints. 

5.  27  da.  18  hr.  49  min.  to  seconds. 

Reduction  Ascending 

Example.  —  Reduce  1306  gills  to  higher  denominations. 

4)1.306  gi. Since  in  1  pt.  there  are  4  gi.,  in  1306  gi. 

2)326  pt.  +  2  gi.  there  are  as  many  pints  as  4  gi.  are  contained 

4)163  qt.. times  in  1306  gi.,  or  .326  pt.  and  2  gi.  remainder. 

40  gal.  +  3  qt.  In  the  same  way  the  quarts  and  gallons  are 

40  gal.  3  qt.  2  gi.  Ans.       found.     So  there  are  in  1306  gi.,  40  gal.  3  qt. 
2gi. 


REVIEW   OF  ARITHMETIC  43 

EXAMPLES 

Reduce  to  higher  denominations : 

1.  Reduce  225,932  in.  to  miles,  etc. 

2.  Change  1384  dry  pints  to  higher  denominations. 

3.  In  139,843  sq.  in.  how  many  square  miles,  rods,  etc.  ? 

4.  How  many  cords  of  wood  in  3692  cu.  ft.  ? 

5.  How  many  bales  in  24,000  sheets  of  paper  ? 

A  denominate  fraction  is  a  fraction  of  a  unit  of  weight  or 
measure. 

To  reduce  denominate  fractions  to  integers  of  lower  denominations. 

Change  the  fraction  to  the  next  lower  denomination.  Treat 
the  fractional  part  of  the  product  in  the  same  ivai/,  and  so  pro- 
ceed to  the  required  denomination. 

Example.  —  Reduce  4  of  a  mile  to  rods,  yards,  feet,  etc. 
^  of  320  rd.  ^  J-V^  rd.  =  228|  rd. 

^of  Vyd.  =  nyd=3Jyd. 

^  f  of  3  ft.  =  Of  ft. 
f  of  12  in.  =  5^  in.  =5f  in. 
f  of  a  mile  =  228  rd.  3  yd.  0  ft.  6f  in. 

The  same  process  applies  to  denominate  decimals. 

To  reduce  denominate  decimals  to  denominate  numbers. 

Example.  —  Reduce  .87  bu.  to  pecks,  quarts,  etc. 

.87  bu.  .84  qt. 

Change  the  decimal  fraction  to 
the  next  lower  denomination.  Treat 
the  decimal  part  of  the  product  in  the 
same  way,  and  so  proceed  to  the  re- 
quired denomination. 


3.48 

pk.            1.68 

.48  pk. 
8 
3.84  qt. 

pt. 

3pk. 

,  3  qt.  1.68  pt. 

Ans. 

44  VOCATIONAL  MATHEMATICS 

EXAMPLES 
Reduce  to  integers  of  lower  denominations  : 

1.  J  of  an  acre.  3.    ^  of  a  ton. 

2.  .3125  of  a  gallon.  4.   .51625  of  a  mile. 

5.  Change  f  of  a  year  to  months  and  days. 

6.  .2364  of  a  ton. 

7.  What  is  the  value  of  ^  of  1|  of  a  mile  ? 

8.  Reduce  |^  bu.  to  integers  of  lower  denominations. 

9.  .375  of  a  month. 

10.   y^y  acre  are  equal  to  how  many  square  rods,  etc.  ? 

Addition  of  Compound  Numbers 

Example.  —  Find  the  sum  of  7  hr.  30  min.  45  sec,  12  hr. 
25  min.  30  sec,  20  hr.  15  min.  33  sec,  10  hr.  27  min.  46  sec. 

The  sum  of  the  seconds  =  154  sec.  = 
2  min.  34  sec.     Write  the  34  sec.  under 
the  sec.  column  and  add  the  2  rain,  to 
the  min.  column.    Add  the  other  columns 
50        39        34  in  the  same  way. 

60  hr.  39  min.  34  sec.   Ans. 

Subtraction  of  Compound  Numbers 

Example.  —  From  39  gal.  2  qt.  2  pt.  1  gi.  take  16  gal.  2  qt. 
3  pt.  3  gi. 

As  3  gi.  cannot  be  taken  from  1  gi.,  4  gi. 
or  1  pt.  are  borrowed  from  the  pt.  column 
and  added  to  the  1  gi.  Subtract  3  gi.  from 
the  5  gi.  and  the  remainder  is  2  gi.  Continue 
in  the  same  way  until  all  are  subtracted. 
Then  the  remainder  is  22  gal.  3  qt.  0  pt.  2  gi. 


hr. 

min. 

sec. 

7 

30 

45 

12 

25 

30 

20 

15 

33 

10 

27 

46 

gal. 

qt. 

pt. 

gi. 

39 

2 

2 

1 

16 

2 

3 

3 

22 

3 

0 

2 

22  gal. 

3qt 

.2 

gi.  Ans. 

yd. 

ft. 

In. 

4 

2 

8 

8 

39 

0 

4 

39  yd 

.4i 

in.  Ans. 

REVIEW  OF  ARITHMETIC  45 

Multiplication  of  Compound  Numbers 
Example.  —  Multiply  4  yd.  2  ft.  8  in.  by  8. 

8  times  8  In.  =  64  in.  =  5  ft.  4  in.  Place  the 
4  in.  under  the  in.  column,  and  add  the  5  ft.  to 
the  product  of  2  ft.  by  8,  which  equals  21  ft.  =  7  yd. 
Add  7  yd.  to  the  product  of  4  yd.  by  8  =  39  yd. 


Division  of  Compound  Numbers 
Example.  —  Fiud  ^  of  42  rd.  4  yd.  2  ft.  8  in. 


-^  of  42  rd.  =  1  rd. ;  re- 
mainder, 7  rd.  =  38^  yd. ; 
add  4  yd.  =  42}  yd.  ^5  of 
42^  yd.  =  1  yd. ;  remainder, 
7i  yd.,  =  22i  ft.  =  24J  ft. 
5^  of  24 J  ft.  =  0  ft.  24 i  ft. 
=294  in.  ;  add  8  in.  =302  in. 
^  of  302  in.  =  8Jf  in. 

Ans. 


Difference  between  Dates 

Example.  —  Find  the  time  from  Jan.  25,  1842,  to  July  4, 
1896. 

1896  7  4  It  is  customary  to  consider  30  days 

1842  1  25 to  a  month.    July  4,  1896,  i.s  the  1896th 

54  yr.     5  mo.    9  da.   Ans.      yr.,  7th  mo.,  4th  da.,  and  Jan.  25,  1842, 

is  the  1842d   yr.,   Ist.   mo.,  25th  da. 

Subtract,  taking  30  da.  for  a  month. 


Id.  yd. 

ft.   in. 

36)42   4 

2   8(1  rd. 

36 

i 

^ 

H 

35)24^(0  ft. 

35 

12 

38i 

294 

+  4 

+  8 

35)42J  yd.  (1  yd.  36)iM(8|^  in. 

35 

280 

7i 

22 

3 

22i  ft. 

1  rd.  1  yd.  Sji  in. 

12 

46  VOCATIONAL  MATHEMATICS 

Example.  —  What  is  the  exact  number  of  days  between 
Dec.  16,  1895,  and  March  12,  1896  ? 

Dec.   15  Do  not  count  the  first  day  mentioned.    Tliere 

Jan.   31  are  15  days  in  December,  after  the  16th.    Jan- 

Feb.  29  uary  has  31  days,  February  29  (leap  year), 

Mar.  12  and  12  days  in  March  ;  making  87  days. 

87  days.  Ans. 

EXAMPLES 

1.  How  much  time  elapsed  from  the  landing  of  the  Pil- 
grims, Dec.  11,  1620,  to  the  Declaration  of  Independence, 
July  4,  1776? 

2.  Washington  was  born  Feb.  22,  1732,  and  died  Dec.  14, 
1799.     How  long  did  he  live? 

3.  Mr.  Smith  gave  a  note  dated  Feb.  25,  1896,  and  paid  it 
July  12, 1896.  Find  the  exact  number  of  days  between  its  date 
and  the  time  of  payment. 

4.  A  carpenter  earning  $2.50  per  day  commenced  Wednes- 
day morning,  April  1, 1896,  and  continued  working  every  week 
day  until  June  6.     How  much  did  he  earn  ? 

5.  Find  the  exact  number  of  days  between  Jan.  10,  1896, 
and  May  5,  1896. 

6.  John  goes  to  bed  at  9.15  p.m.  and  gets  up  at  7.10  a.m. 
How  many  minutes  does  he  spend  in  bed  ? 

To  multiply  or  divide  a  compound  number  by  a  fraction. 

To  multiply  by  a  fraction,  multiply  by  the  numerator,  and 
divide  the  product  by  the  d^nomiriator. 

To  divide  by  a  fraction,  multiply  by  the  denominator ,  and  divide 
the  product  by  the  numerator. 

When  the  multiplier  or  divisor  is  a  mixed  number,  reduce  to 
an  improper  fraction,  and  proceed  as  above. 


REVIEW  OF  ARITHMETIC  47 

EXAMPLES 

1.  How  much  is  4  of  16  hr.  17  min.  14  sec.  ? 

2.  A  field  contains  10  acres  12  sq.  rd.  of  lan«l,  which  is  J  of 
the  whole  farm.     Find  the  size  of  the  farm. 

3.  If  a  train  runs  60  mi.  35  rd.  16  ft.  in  one  hour,  how  far 
will  it  run  in  12f  hr.  at  the  same  rate  of  speed  ? 

4.  Divide  14  bu.  3  pk.  6  qt.  1  pt.  by  J. 

5.  Divide  5  yr.  1  mo.  1  wk.  1  da.  1  hr.  1  min.  1  sec.  by  3J. 

EXAMPLES 

1.  A  time  card  on  a  piece  of  work  states  that  2  hours  and 
15  minutes  were  spent  in  lathe  work,  1  hour  and  12  minutes  in 
milling,  2  hours  and  45  minutes  in  planing,  and  1  hour  and  30 
minutes  on  bench  work.  What  was  the  number  of  hours  spent 
on  the  job  ? 

2.  How  many  castings,  each  weighing  14  oz.,  can  be 
obtained  from  860  lb.  of  metal  if  nothing  is  allowed  for 
waste? 

3.  How  many  feet  hmg  must  a  machine  shop  be  to  hold  a 
lathe  8'  6",  a  planer  14'  4",  a  milling  machine  4'  2",  and  a 
lathe  7'  5",  placed  side  by  side?  3'  3"  were  allowed  between 
the  machines  and  between  the  walls  and  the  machines. 

4.  How  many  gross  in  a  lot  of  968  screws  ? 

5.  Find  the  sum  of  7  hr.  30  min.  45  sec,  12  hr.  26  min. 
30  sec,  20  hr.  15  min.  33  sec,   10  hr.  27  min.  46  sec. 

6.  If  a  train  is  run  for  8  hours  at  the  average  rate  of 
50  mi.  30  rd.  10  ft.  per  hour,  how  great  is  the  distance  covered? 

7.  A  telephone  pole  is  31  feet  long.  If  4  ft.  7  in.  are 
under  ground,  how  high  (in  inches)  is  the  top  of  the  pole  above 
the  street? 

8.  If  100  bars  of  iron,  each  2}'  long,  weigh  70  lb.,  what  is 
the  total  weight  of  2300  ? 


48  VOCATIONAL  MATHEMATICS 

9.    If  43  in.  are  cut  from  a  wire  3  yd.  2  ft.  6  in.  long,  what 
is  the  length  of  the  remaining  piece  ? 

10.  If  a  rod  of  iron  18'  8"  long  is  cut  into  pieces  6J"  long 
and  Jj"  is  allowed  for  waste  in  each  cut,  how  many  pieces 
can  be  cut  ?     How  much  remains  ? 

11.  I  have  84  lb.  14  oz.  of  salt  which  I  wish  to  put  into 
packages  of  2  lb.  6  oz.  each.     How  many  packages  will  there  be  ? 

12.  If  one  bottle  holds  1  pt.  3  gi.,  how  many  dozen  bottles 
will  be  required  to  hold  65  gal.  2  qt.  1  pt.  ? 

13.  How  many  pieces  5J"  long  can  be  cut  from  a  rod  16'  8" 
long,  if  5"  is  allowed  for  waste  ? 

14.  What  is  the  entire  length  of  a  railway  consisting  of 
five  different  lines  measuring  respectively  160  mi.  185  rd.  2  yd., 
97  mi.  63  rd.  4  yd.,  126  mi.  272  rd.  3  yd.,  67  mi.  199  rd.  5  yd., 
and  48  mi.  266  rd.  5  yd.  ? 

Percentage 

Percentage  is  a  process  of  solving  questions  of  relation  by 
means  of  hundredths  or  per  cent  (%). 

Every  question  in  percentage  involves  three  elements :  the 
rate  per  cent,  the  base,  and  the  percentage. 

The  rate  per  cent  is  the  number  of  hundredths  taken. 

The  base  is  the  number  of  which  the  hundredths  are 
taken. 

The  percentage  is  the  result  obtained  by  taking  a  certain  per 
cent  of  a  number. 

Since  the  percentage  is  the  result  obtained  by  taking  a  cer- 
tain per  cent  of  a  number  it  follows  that,  the  percentage  is  the 
product  of  the  base  and  the  rate.  The  rate  and  base  are  always 
factors,  the  percentage  is  the  product. 

Example.  —  How  much  is  8  %  of  $200? 

8%  of  $  200  =  200  X  .08  =  $  16.  (1) 


REVIEW  OF  ARITHMETIC  49 

In  (1)  we  have  the  three  elements:  8%  is  the  rate,  8200  is  the  base, 
and  $  16  is  the  percentage. 

Since  $200  x  .08  =  $  16,  the  percentage  ; 

$  16  -i-  .08  =  $  200,  the  base  ; 
and  $  16  -i-  $  200  =  .08,  the  rate. 

If  any  two  of  these  elements  are  given,  the  other  may  be 
found : 

Base  X  Rate  =  Percentage 
Percentage  -5-  Rate  =  Base 
Percentage  -5-  Base  =  Rate 

Per  cent  is  commonly  used  in  the  decimal  form,  but  many 
operations  may  be  much  shortened  by  using  the  common  frac- 
tion form. 

1  %  =    .01  =  ^^  i%=  .00^  or  .005 

10%=    .10  =  ^V  33J%  =  .33J  =  i 
100  %  =  1.00  =  1  8^  %  =  .08^  =  .0825 

12^  %  =  .12^  or  .125  =  ^  |  %  =  .00^  =  .00125 

There  are  certain  per  cents  that  are  used  so  frequently  that 
we  should  memorize  their  equivalent  fractions. 


10%=^  37-^%=!  75%  =  I 

12i%=i  40%  =  J  80%=  J 

16i%=i  50%  =i  83i%=J 

20%  =i  60%=  I  87i%=J 

EXAMPLES 

1.  Find  75  %  of  $  368. 

2.  Find  15%  of  S412. 

3.  840  is  33  J  %  of  what  number  ? 

4.  615  is  15%  of  what  number? 

5.  What  per  cent  of  12  is  8  ? 


50  VOCATIONAL  MATHEMATICS 

6.  What  per  cent  of  245  is  5  ? 

7.  What  per  cent  of  195  is  39  ? 

8.  What  per  cent  of  640  is  80  ? 

9.  What  per  cent  of  750  is  25  ? 
10.  What  per  cent  of  819  is  45  ? 

Trade  Discount 

Merchants  and  jobbers  have  a  price  list.  From  this  list 
they  give  special  discounts  according  to  the  credit  of  the 
customer  and  the  amount  of  supplies  purchased,  etc.  If  they 
give  more  than  one  discount  it  is  understood  that  the  first 
means  the  discount  from  the  list  price,  the  second  denotes  the 
discount  from  the  remainder. 

EXAMPLES 

1.  What  is  the  price  of  200  No.  1  cleats  at  $  36.68  per  M. 
at  40%  off? 

2.  Supplies  from  a  hardware  store  amounted  to  $  58,75.  If 
121  %  were  allowed  for  discount,  what  was  the  amount  paid  ? 

3.  A  dealer  received  a  bill  amounting  to  $  212.75.  Succes- 
sive discounts  of  75%,  15%,  10%,  and  5%  were  allowed. 
What  was  amount  to  be  paid  ? 

4.  2  %  is  usually  discounted  on  bills  paid  within  30  days. 
If  the  following  are  paid  within  30  days,  what  will  be  the 
amounts  due? 

a.  $2816.49  d.    $1369.99  g.    $4916.01 

b.  399.16  e.      2717.02  h.  30.19 

c.  489.01  /.        918.69 

5.  What  is  the  price  of  20  fuse  plugs  at  $  .07  each,  30  % 
off? 

6.  Hardware  supplies  amounted  to  $  127.79  with  a  discount 
of  40  and  15  %.     What  was  the  net  price? 


REVIEW  OF  ARITHMETIC  51 

7.  Which  is  better,  for  a  merchant  to  receive  a  straight  dis- 
count of  95  %  or  a  successive  discount  of  75,  15,  5  %  ? 

8.  Twenty  per  cent  is  added  to  the  number  of  workmen  in 
a  machine  shop  of  575  men.  What  is  the  number  employed 
after  the  increase  ? 

9.  A  steam  pressure  of  180  lb.  per  square  inch  is  raised 
to  225  lb.  per  square  inch.  What  is  the  per  cent  of  in- 
crease ? 

10.  If  31  out  of  595  wheels  are  rejected  because  of  defects, 
what  per  cent  is  rejected  ? 

11.  A  clerk's  salary  was  increased  OJ  %.  If  he  now  receives 
$  850,  what  was  his  original  salary  ? 

12.  A  company  lost  12|^  %  of  its  men  and  had  560  left. 
How  many  men  were  there  before  ? 

Simple  Interest 

Money  that  is  paid  for  the  use  of  money  is  called  interest. 
The  money  for  the  use  of  which  interest  is  paid  is  called  the 
principal^  and  the  sum  of  the  principal  and  interest  is  called  the 
amojint. 

Interest  at  6  %  means  6%  of  the  principal  for  1  year;  12 
months  of  30  days  each  are  usually  regarded  as  a  year  in  com- 
puting interest. 

Example.  —  What  is  the  interest  on  $  100  for  3 years  at  6  %  ? 

8100 
.06 


$ 6.00  interest  for  one  year.  Or,  yg^  x  Jf *i  x  f  =  $  18.     Ans. 

3 


%  18.00  interest  for  3  years.   Ana. 

8 100  -I-  §  18  =  $  118,  amount. 
Pnncfpal  X  Kate  x  Time  =  Literest. 


52  VOCATIONAL  MATHEMATICS 

Example.  —  What  is  the  interest  on  $  297.62  for  5  yr.  3  mo 
at6%? 

1297.62 

ni:sM         o,,  1_  ^  $  297.62     21  ^  118750.06  ^    ,3^, 
.,  100  1  ^  200 


44643 

892860  Note. — Final    results    should    not    include   mills. 

$93.7503  Mills  are  disregarded  if  less  than  5,  and  called  another 

$  93.75.  Ans.  cent  if  5  or  more. 

EXAMPLES 

1.  What  is  the  interest  on  $  586.24  for  3  months  at  6  %  ? 

2.  What  is  the  interest  on  $  816.01  for  9  months  at  5  %  ? 

3.  What  is  the  interest  on  $  314.72  for  1  year  at  4  %  ? 

4.  What  is  the  interest  on  $  876.79  for  2  yr.  3  mo.  at  41  fo  ? 


The  Six  Per  Cent  Method 

By  the  6  %  method  it  is  convenient  to  find  first  the  interest 
of  $  1,  then  multiply  it  by  the  principal. 

Example.  —  What  is  the  interest  of  $  50.24  at  6  %  for  2  yr. 
8  mo.  18  da.  ? 

Interest  on  $;  1  for  2  yr.  =2  x  $  .06  =  .$  .12 
Interest  on  §  1  for  8  mo.  =8  x  $  .00^  =  .04 
Interest  on  $  1  for  18  da.  =  18  x  $  .000^  =  .003 
Interest  on  $  1  for  2  yr.  8  mo.  18  da.  .$  .163 

Interest  on  ^50.24  is  50.24  times  $  .163  =  .$8.19.    Ans. 
Find  the  interest  on  $  1  for  the  given  time,  and  multiply  it  by  the  prin- 
cipal, considered  as  an  abstract  number. 

EXAMPLES 
Find  the  interest  and  amount  of  the  following : 

1.  $  2350  for  1  yr.  3  mo.  6  da.  at  5  %. 

2.  $  125.75  for  2  mo.  18  da.  at  7  %. 

3.  $  950.63  for  3  yr.  17  da.  at  41  %. 

4.  $  625.57  for  1  yr.  2  mo.  15  da.  at  6  %. 


REVIEW   OF  ARITHMETIC  53 

Exact  Interest 

When  the  time  includes  clays,  interest  computed  by  the  6% 
method  is  not  strictly  exact,  by  reason  of  using  only  .'^0  days 
for  a  month,  which  makes  the  year  only  360  days.  The  day  is 
therefore  reckoned  as  -^  of  a  year,  whereas  it  is  ^^-y  of  a  year. 

To  compute  exact  interest^  find  the  exact  time  in  daya^  and  coiir 
aider  1  day^s  interest  as  ^^  of  1  yea^s  interest. 

Example.  —  Find  the  exact  interest  of  $  358  for  74  days  at 
7  %. 

$:368  X  .07  =  $26.06,  1  year's  interest. 
74  days*  interest  is  j\^  of  1  year's  interest. 
3Vir  of  $26.06  =  $6.08.   Ans. 

'      1         100      366 

EXAMPLES 
Find  the  exact  interest  of : 

1.  $324  for  15  da.  at  5  %. 

2.  $253  for  98  da.  at  4%. 

3.  $624  for  117  da.  at  7%. 

4.  $  620  from  Aug.  15  to  Nov.  12  at  6  %. 

5.  $  153.26  for  256  da.  at  5^  %. 

6.  S  540.25  from  June  12  to  Sept.  14  at  8  %. 

Rules  for  Computing  Interest 

The  following  will  be  found  to  be  excellent  rules  for  finding  the  inter* 
est  on  any  principal  for  any  number  of  days. 

Divide  the  principal  by  100  and  proceed  as  follows: 

2  %  — Multiply  by  number  of  days  to  run,  and  divide  by  180. 
2^  %  —  Multiply  by  number  of  days,  and  divide  by  144. 

3  %  —  Multiply  by  number  of  days,  and  divide  by  120. 

3^  %  —  Multiply  by  number  of  days,  and  divide  by  102.86. 


54 


VOCATIONAL  MATHEMATICS 


4  %  —  Multiply  by  number  of  days,  and  divide  by  90. 

5  %  —  Multiply  by  number  of  days,  and  divide  by  72. 

6  %  —  Multiply  by  number  of  days,  and  divide  by  60. 

7  %  —  Multiply  by  number  of  days,  and  divide  by  51.43. 

8  %  — Multiply  by  number  of  days,  and  divide  by  45. 


Savings  Bank  Compound  Interest  Table 

Showing  the  amount  of  $  1,  from  1  year  to  15  years,  with  compound 
interest  added  semiannually,  at  different  rates. 


Pkb  Cent 

3 

4 

5 

6 

7 

8 

9 

h  year 

101 

102 

102 

103 

103 

1  04 

104 

1    year 

103 

104 

105 

106 

107 

108 

109 

1^  yeare 

104 

106 

107 

109 

110 

1  12 

1  14 

2    yeai*s 

106 

108 

1  10 

1  12 

1  14 

116 

119 

2^  years 

107 

1  10 

113 

1  16 

1  18 

1  21 

124 

3    years 

109 

1  12 

1  15 

1  19 

1  22 

1  26 

130 

3^  years 

110 

1  14 

118 

122 

127 

131 

136 

4    years 

1  12 

1  17 

121 

126 

131 

1  36 

142 

4|  years 

1  14 

1  19 

124 

130 

136 

142 

148 

5    years 

1  16 

121 

128 

134 

141 

148 

165 

5^  years 

1  17 

124 

131 

138 

145 

163 

162 

6    years 

1  19 

126 

134 

142 

1  61 

160 

169 

6^  years 

121 

129 

137 

146 

156 

166 

1  77 

7    years 

123 

131 

141 

1  51 

161 

1  73 

185 

7^  years 

124 

1  34 

144 

166 

167 

180 

193 

8    years 

126 

137 

148 

1  60 

173 

187 

2  02 

8|  years 

128 

139 

152 

166 

179 

194 

2  11 

9    years 

1  30 

142 

1  66 

170 

185 

2  02 

2  20 

9^  years 

132 

146 

1  59 

176 

192 

2  10 

2  30 

10    years 

1  34 

148 

163 

180 

1  98 

2  19 

2  41 

11    years 

138 

1  54 

1  72 

191 

2  13 

2  36 

2  63 

12    years 

142 

160 

180 

2  03 

2  28 

2  56 

2  87 

13    years 

147 

167 

1  90 

2  16 

2  44 

2  77 

3  14 

14    years 

1  61 

1  73 

1  99 

2  28 

2  62 

2  99 

3  42 

16    years 

1  66 

1  80 

2  09 

2  42 

2  80 

3  24 

3  74 

REVIEW   OF  ARITHMETIC 


EXAMPLES 


Solve  the  following  problems  according  to  the  tables  given 
above : 

1.  What  is  the  compound  interest  of    $  1  at  the   end  of 
8.1  years? 

2.  What  is  the  compound  interest  of  $1  at  the  end  of  11 
years  ? 

3.  How  long  will   it  take    $400  to  double  itself  at  4%, 
compound  interest  ? 

4.  How  long  will  it  take  $580  to  double  itself  at  4^  %, 
compound  interest? 

5.  How  long  will  it  take  $615  to  double  itself  at  6%, 
simple  interest  ? 

6.  How  long  will  it  take  $784  to  double  itself  at  5|  %, 
simple  interest  ? 

7.  Find  the  interest  of  S684  for  94  days  at  3  %. 
a    Find  the  interest  of  $  1217  for  37  days  at  4  %. 

9.    Find  the  interest  of  $681.14  for  74  days  at  4^  %. 

10.  Find  the  interest  of  $414.50  for  65  days  at  5  %. 

11.  Find  the  interest  of  $384.79  for  115  days  at  6  %. 

Ratio  and  Proportion 

Ratio  is  the  relation  between  two  numbers.  It  is  found 
by  dividing  one  by  the  other.     The  ratio  of  4  to  8  is  4  -§-  8  =  i. 

The  terms  of  the  ratio  are  the  two  numbers  compared.  The 
first  term  of  a  ratio  is  the  antecedent^  and  the  second  the  con- 
sequent. The  sign  of  the  ratio  is  (:).  (It  is  the  division  sign 
with  the  line  omitted.)  Ratio  may  also  be  expressed  fraction- 
ally, as  J/  or  16:4;  or  yV  or  3  :  17. 

A  ratio  formed  by  dividing  the  consequent  by  the  antece- 
dent is  an  inverse  raJtio :  12  :  6  is  the  inverse  ratio  of  6  :  12. 


56  VOCATIONAL  MATHEMATICS 

The  two  terms  of  the  ratio  taken  together  form  a  couplet. 
Two  or  more  couplets  taken  together  form  a  compound  ratio. 
Thus,  *  3  :  6  =  23  :  46 

A  compound  ratio  may  be  changed  to  a  simple  ratio  by  tak- 
ing the  product  of  the  antecedents  for  a  new  antecedent,  and 
the  product  of  the  consequents  for  a  new  consequent. 

Antecedent  -i-  Consequetit  =  Ratio 

Antecedent  -;-  Ratio  =  Consequent 
Ratio  X  Consequent  =  Antecedent 

To  multiply  or  divide  both  terms  of  a  ratio  by  the  same 
number  does  not  change  the  ratio. 

Thus  12 : 6  =  2 

3x12:3x6  =  2 

EXAMPLES 
Find  the  ratio  of 

1.  20  :  300  Fractions  with  a  common  de- 

2.  3  bu. :  3  pk.  nominator  have  the  same 
3    21-16  ratio  as  their  numerators. 

4.     12  :  4-  7         8_.16     28.7       15.30 

•     -^*^  •  4  '•     IT  •  IT?  TT-  7  5J    11  •  TT 

5-    i  =  l  a    |:|,B:|,.2.5 

6.    16:(?)  =  | 

Proportion 

An  equality  of  ratios  is  a  proportion. 

A  proportion  is  usually  expressed  thus :  4  :  2  :  :  12  :  6,  and  is 
read  4  is  to  2  as  12  is  to  6. 

A  proportion  has  four  terms,  of  which  the  first  and  third  are 
antecedents  and  the  second  and  fourth  are  consequents.  The 
first  and  fourth  terms  are  called  extremes,  and  the  second  and 
third  terms  are  called  means. 

The  product  of  the  extremes  equals  the  product  of  the 
means. 


REVIEW  OP  ARITHMETIC  57 

Tojind  an  extreme,  divide  the  product  of  the  means  Ini  the  given 
extreme. 

Tojind  a  mean,  divide  the  proilact  of  the  extremes  by  the  given 
mean. 

EXAMPLES 

Supply  the  missing  term  : 

1.  1:836::25:()  4.    10  yd.  :  50  yd.  : :  $  20  :  ($   ) 

2.  6:24::(  )  :  40  5.    $}:S3|::(  ):5 

3.  (  ):15::60:6 

Simple  Proportion 

An  equality  of  two  simple  ratios  is  a  simple  proportion. 
Example.  —  If  12  bushels  of  charcoal  cost  $4,  what  will  60 
bushels  cost  ? 

12  •  60  •  •  S4  •  f.«5  ^  There  is  the  same  relation  between  the  cost 

„  ■     , "       '^''  ''  of  12  bu.  and  the  cost  of  60  bu.  as  there  is  be- 

— -|— =$20.     Am.      tween  the  12  bu.  and  the  60  bu.      $4  is  the 

third  term.     The  answer  is  the  fourth  term. 

It  must  form  a  ratio  of  12  and  60  that  shall  equal  the  ratio  of  §  4  to  the 

answer.     Since  the  third  term  is  lass  than  the  required  answer,  the  first 

must  be  less  than  the  second,  and  12  :  60  is  the  first  ratio.     The  product 

of  the  means  divided  by  the  given  extreme  gives  the  other  extreme,  or  $  20. 

EXAMPLES 
Solve  by  proportion : 

1.  If  150  fuses  cost  $  6,  how  much  will  1200  cost  ? 

2.  If  250  pounds  of  lead  pipe  cost  S 15,  how  much  will  1200 
pounds  cost  ? 

3.  If  5  men  can  dig  a  ditch  in  3  days,  how  long  will  it  take 
2  men  ? 

4.  If  .4' men  can  shingle  a  shed  in  2  days,  how  long  will  it 
take  3  men  ? 

5.  The  ratio  of  Simon's, pay  to  Matthew^s  is  |.      Simon 
earns  $  18  per  week.     What  does  Matthew  earn  ? 


58  VOCATIONAL  MATHEMATICS 

6.  What  will  llf  yards  of  cambric  cost  if  50  yards  cost 
$  6.75  ? 

7.  A  spur  gear  making  210  revolutions  per  minute  is  en- 
meshed with  a  pinion.  The  gear  has  126  teeth  and  the  pinion 
has  42  teeth.     How  many  revolutions  does  the  pinion  make  ? 

8.  In  a  velocity  diagram  a  line  3|"  long  represents  45  ft. 
What  would  be  the  length  of  a  line  representing  30  ft.  velocity  ? 

9.  How  many  pounds  of  lead  and  tin  would  it  take  to  make 
4100  pounds  of  solder  if  there  are  27  pounds  of  tin  in  each  100 
pounds  of  solder  ? 

10.  It  is  necessary  to  obtain  a  speed  reduction  of  7  to  3  by 
use  of  gears.  If  the  pinion  has  21  teeth,  how  many  teeth 
must  the  gear  have  ? 

11.  A  bar  of  iron  3^  ft.  long  and  |"  diameter  weighs  6.64 
pounds.  W^hat  would  a  bar  4|^  ft.  long  of  the  same  diameter 
weigh  ? 

12.  In  a  certain  time  15  workmen  made  525  pulleys.  How 
many  pulleys  will  32  men  make  in  the  same  length  of  time  ? 

13.  When  a  post  11.5  ft.  high  casts  a  shadow  on  level 
ground  20.6  ft.  long,  a  telephone  pole  near  by  casts  a  shadow 
59.2  ft.  long.     How  high  is  the  pole  ? 

14.  The  diameter  of  a  driving  pulley  is  18".  This  pulley 
makes  320  revolutions  per  minute.  What  must  be  the  diam- 
eter of  a  driving  pulley  in  order  to  make  420  revolutions  per 
minute  ? 

15.  A  ditch  is  dug  in  14  days  of  8  hours  each.  How  many 
days  of  10  hours  each  would  it  have  taken  ? 

16.  If  in  a  drawing  a  tree  38  ft.  high  is  represented  by  IJ", 
what  on  the  same  scale  will  represent  the  height  of  a  house  47 
ft.  high? 

17.  What  will  be  the  cost  of  21  motors  if  15  motors  cost 
$  887.509  ? 


REVIEW  OF  ARITHMETIC  59 

18.  The  main  drive  pulley  of  a  machine  is  6  inches  in  diam- 
eter and  makes  75G  revolutions  per  minute.  A  pulley  on  the 
line  shaft  is  belted  to  a  machine.  What  is  the  diameter  of  the 
line  shaft  pulley  if  the  line  shaft  makes  252  revolutions  per 
minute  ? 

19.  If  a  pole  8  ft.  high  casts  a  shadow  4 J  ft.  long,  how  high 
is  a  tree  which  casts  a  shadow  48  ft.  long  ? 

Involution 

The  product  of  equal  factors  is  a  power. 

The  process  of  finding  powers  is  involution. 

The  product  of  two  equal  factors  is  the  second  power,  or 
square,  of  the  equal  factor. 

The  product  of  three  equal  factors  is  the  third  power,  or  cube, 
of  the  factor. 

4^  =  4  X  4  is  4  to  the  second  power,  or  the  square  of  4. 

2*  =  2  X  2  X  2  is  2  to  the  third  power,  or  the  cube  of  2. 

3^  =  3x3x3x3  is  3  to  the  fourth  power,  or  the  fourth  power  of  4. 

EXAMPLES 


Find  the  powers : 

1.    5»                       3.    1^                        5.    (2^)2 

7.    9^ 

2.   1.1'                 4.   2o'                  6.   2* 

a  .15« 

Evolution 

One  of  the  eqiuil  factors  of  a  power  is  a  root. 

One  of  two  equal  factors  of  a  number  is  the  square  root. 

One  of  three  equal  factors  of  a  number  is  the  cube  root  of  it. 

The  square  root  of  16  =  4.     The  cube  root  of  27  =  3, 

The  radical  sign  (^)  placed  before  a  number  indicates  that 
its  root  is  to  be  found.  The  radical  sign  alone  before  a  number 
indicates  the  square  root. 

Thus,  Vd  =  3  is  read,  the  square  root  of  9  =  3. 


60  VOCATIONAL  MATHEMATICS 

A  small  figure  placed  in  the  opening  of  the  radical  sign  is 
called  the  index  of  the  root,  and  shows  what  root  is  to  be 
taken. 

Thus,  -v^  =  2  is  read,  the  cube  root  of  8  is  2. 

Square  Root 

The  square  of  a  number  composed  of  tens  and  units  is  equal 
to  the  square  of  the  tens,  plus  twice  the  product  of  the  tens  by 
the  units,  plus  the  square  of  the  units. 

ten^  +  2  X  te7is  X  units  +  U7iits^ 

Example.  —  What  is  the  square  root  of  1225? 

12^25(30  +  5^35  Separating 

Tens%  302  =       900  into  periods  of 


325  two     figures 

325  each,     by     a 


2  X  tens  =  2  x  30  =60 

2  X  tens  +  units  =  2  x  30  +  6  =  65 

checkmark  ('), 
beginning  at  units,  we  have  12 '25.  Since  there  are  two  periods  in  tlie 
power,  there  must  be  two  figures  in  the  root,  tens  and  units. 

The  greatest  square  of  even  tens  contained  in  1225  is  900,  and  its 
square  root  is  30  (3  tens).  Subtracting  the  square  of  the  tens,  900,  the 
remainder  consists  of  2  x  (tens  x  units)  +  units. 

325,  therefore,  is  composed  of  two  factors,  units  being  one  of  them, 
and  2  x  tens  —  units  being  the  other.  But  the  greater  part  of  this  factor 
is  2  X  tens  (2  x  30  =  60).  By  trial  we  divide  325  by  60  to  find  the  other 
factor  (units),  which  is  5,  if  correct.  Completing  the  factor,  we  have 
2  X  tens  +  units  =;  66,  which,  multiplied  by  the  other  factor,  5,  gives  325. 
Therefore  the  square  root  is  30  +  5  =  35. 

The  area  of  every  square  surface  is  the  product  of  two  equal 
factors,  length,  and  width. 

Finding  the  square  root  of  a  number,  therefore,  is  equivalent 
to  finding  the  length  of  one  side  of  a  square  surface,  its  area 
being  given. 

1.  Length  X  Width  —Area 

2.  Area     -r-  Length  =  Width 

3.  Area    -^•  Width  =  Jjength    , 


REVIEW  OP  ARITHMETIC  61 

Short  Mkthod 

Example.  —  Find  the  square  root  of  1300.0990. 

13'0«.(W9<5 (8rtJ4  BeginninK  at  the  decimal  point,  separate  the 

9  number  into  i)eri<HlH  of  two  figurea  each,  point- 

06)  400  ing  whole  numbers  to  the  left  and  decimalri  to 

3iHJ  the  right.     Kind  the  greatest  square  in  the  left- 

721)1009  hand  period,  and  write  its  root  at  the  right. 

721  Subtract  the  scjuare  from  the  left-hand  period, 

7224)28890  and  bring  down  the  next  period  for  a  dividend. 

28896  Divide    the  dividend,   with  its   right-hand 

figure  omitted,  by  twice  the  root  already  found, 

and  annex  the  quotient  to  the  root,  and  to  the  divisor.     Multiply  thlH 

complete  divisor  by  the  last  root  figure,  and  bring  down  the  next  period 

for  a  dividend,  as  before. 

Proceed  in  this  manner  till  all  the  periods  are  exhausted. 
When  0  occurs  in  the  root,  annex  0  to  the  trial  divisor,  bring  down 
the  next  period,  and  divide  as  before. 

If  there  is  a  remainder  after  all  the  periods  are  exhausted,  annex  deci- 
mal periods. 

If,  after  multiplying  by  any  root  figure,  the  product  is  larger  than  the 
dividend,  the  root  figure  is  too  large  and  must  be  diminished.  Also  the 
last  figure  in  the  complete  divisor  must  be  diminished. 

For  every  decimal  period  in  the  power,  there  must  be  a  decimal  figure 
in  the  root.  If  the  last  decimal  period  does  not  contain  two  figures, 
supply  the  deficiency  by  annexing  a  cipher. 

EXAMPLES 
Find  the  square  root  of : 


1.  8830  5.    \l^l  9.    V3.532h-0.28 

2.  370881  6.    72.5  10.    Ve26  + 1290 

3.  29.0521  7.    .009yV  11.    — .  x  ^ 


4.   40050  8.    1684.298431  12. 


V9       '^ 
3909 


5025 

13.    What  is  the  length  of  one  side  of  a  square  field  that  has 
an  area  equal  to  a  field  75  rd.  Ion?:  and  45  rd.  wide  ? 


•  cnmfer 


CHAPTER   II 
MENSURATION 

The  Circle 

A  circle  is  a  plane  figure  bounded  by  a  curved  line,  called 
the  circumference,  every  point  of  which  is  equidistant  from  the 
center. 

The  diameter  is  a  straight  line  drawn 
from  one  point  of  the  circumference 
to  another  and  passing  through  the 
center. 

The  ratio  of  the  circumference  to 
the  diameter  of  any  circle  is  always  a 
constant  number,  3.1416+,  approxi- 
mately 3^,  which  is  represented  by 
the  Greek  letter  tt  (pi). 

C  =  Circumference 
D  =  Diameter 


The  radius  is  a  straight  line  drawn  from  the  center  to  the 
circumference. 

Any  portion  of  the  circumference  is  an  arc. 

By  drawing  a  number  of  radii  a  circle  may  be  cut  into  a 
series  of  figures,  each  one  of  which  is  called  a  sector.  The  area 
of  each  sector  is  equal  to  one  half  the  product  of  the  arc  and 
radius.  Therefore  the  area  of  the  circle  is  equal  to  one  half  of 
the  product  of  the  circumference  and  radius. 

1  See  Appendix  for  use  of  formulas. 
62 


MENSURATION  63 

•  2 

In  this   formula  A   equals   area,  tt  =  3.1410,  and   /?  =  the 
radius  squared. 

^  =  i  Z>  X  i  C 

In  this  formula  D  equals  the  diameter  and  C  the  circum- 
ference, 

4  4 

Example. — What  is  the  area  of   a  circle  whose  radius  is 
3ft.? 

4 


^=irx9        ^=^^  =  ir9  =  28.278q.  ft.     Ans. 
i 

Example.  — What  is  the  area  of  a  circle  whose  circumfer- 
ence is  10  ft.  ? 

2)  =  -^        A=Idx-C 
3.1416  2         2 

-  X  -^~  X  1  X  10  =  -^^  =  7.1  sq.  ft.     Ans. 
2      3.141«      2  3.1410 

Area  of  a  Ring.  —  On  examining  a  flat  iron  ring  it  is  clear  that 

the  area  of  one  side  of  the  ring  may  be  found  by  subtracting 

the  area  of  the  inside  circle  from  the  area  of  the  outside  circle. 

Let  D  =  outside  diameter 

d  =  inside  diameter 

A  —  area  of  outside  circle 

a  =  area  of  inside  circle 

(1)  ^  =  :^=.7854Z>* 

4 


64 


VOCATIONAL    MATHEMATICS 


(2) 


a  =  ^  =  .7854(^2 


(3)    A-a  =  ^~^ 

Let  B  =  area  of  circular  ring  =  A  —  a 

^  =  —  -  ^  =  -  (Z>2  -  d2w  .7854  (Z>2  _  (^2) 
4         4       4  ^ 

Example.  —  If  the  outside  diameter  of  a  flat  ring  is  9"  and 
the  inside  diameter  7",  what  is  the  area  of  one  side  of  the 
ring? 

^=.7854  (D2  _(f2) 

B  =  .7854  (81  -  49)  =  .7854  x  32  =  25.1328  sq.  in.     Ans. 


Angles 

Mechanics  make  two  uses  of  angles  :  (1)  to  measure  a  cir- 
cular movement,  and  (2)  to  measure  a  difference  in  direction. 
A  circle  contains  360°,  and  the  angles  at  the  center  of  the 
circle  contain  as  many  degrees  as  their  corresponding  arcs  on 
the  circumference. 

Angle  FOE  has  as  many  degrees  as  arc  PE. 

A  right  angle  is  measured  by  a  quarter 
of  the  circumference  of  the  circle,  which     ^ 
is  90°. 

The  angle  AOG  is  a  right  angle. 

The  angle  AC,  made  with  half  the  cir- 
cumference of  the  circle,  is  a  straight  angle,  and  the  two  right 
angles,  AOG  and  GOC,  which  it  contains,  are  supplementary 
to  each  other.  When  the  sum  of  two  angles  is  equal  to  90°, 
they  are  said  to  be  complementary  angles,  and  one  is  the  com- 
plement of  the  other.  When  the  sum  of  two  angles  equals  180°, 
they  are  supplementary  angles,  and  one  is  said  to  be  the  supple- 
ment of  the  other. 


MENSURATION 


65 


The  number  of  degrees  in  an  angle  may  be  measured  by  a 
protractor.     The  distance  around  a  semicircular  protractor  is 


Protractor  —  Semicircular,  having  180°. 


divided  into  180  parts,  each  division  measuring  a  degree.  It 
is  used  by  placing  the  center  of  the  protractor  on  the  vertex 
and  the  base  of  the  protractor  on  the  side  of  the  angle  to  be 


3G0°  Protractor. 
A  a  is  a  circle  divided  into  degrees. 

measured.     Where  the  other  side  of  the  angle  cuts  the  circular 
piece,  the  size  of  the  angle  may  be  read. 

EXAMPLES 

1.  What   is   the   area   of   a   circular   sheet   of  iron   8"  in 
diameter  ? 

2.  What  is  the  distance  around  the  edge  of  a  pulley  6"  in 
diameter  ? 


66  VOCATIONAL    MATHEMATICS 

3.  What  is  the  area  of  one  side  of  a  flat  iron  ring  14"  inside 
diameter  and  18"  outside  diameter  ? 

4.  A  driving  wheel  of  a  locomotive  has  a  wheel  center  of 
56"  in  diameter ;  if  the  tires  are  3"  thick,  what  is  the  circum- 
ference of  the  wheel  when  finished  ? 

5.  Find  the  area  of  a  section  of  an  iron  pipe  which  has  an 
inside  diameter  of  17"  and  an  outside  diameter  of  17|". 

6.  Name  the  complements  of  angles  of  30°,  45°,  65°,  70°, 
85°. 

7.  Name  the  supplements  of  angles  of  55°,  140°,  69°,  98°  44', 
81°  19'. 

8.  What  is  the  diameter  of  a  wheel  that  is  12'  6"  in  circum- 
ference ? 

Triangles 

A  triangle  is  a  plane  figure  bounded  by  three  straight  lines. 
Triangles  are  classified  according  to  the  relative  lengths  of 
their  sides  and  the  size  of  their  angles. 

A  triangle  having  equal  sides  is  called  equilateral.  One 
having  two  sides  equal  is  isosceles.  A  triangle  having  no 
sides  equal  is  called  scalene. 

If  the  angles  of  a  triangle  are  equal,  the  triangle  is  equi- 
angular. 

If  one  of  the  angles  of  a  triangle  is  a  right  angle,  the  tri- 
angle is  a  right  triangle.  In  a  right  triangle  the  side  opposite 
the  right  angle  is  called  the  hypotenuse  and  is  the  longest  side. 
The  other  two  sides  of  the  right  triangle  are  the  legs,  and  are 
at  right  angles  to  each  other. 


Equilateral       Isosceles  Scalene  Right 


MENSURATION 


67 


Kinds  of  Triangles 
Right  Triangles 


In  a  right  triangle  the 
square  of  the  hypotenuse 
equals  the  sum  of  the 
squares  of  the  other  two 
sides  or  legs. 

If  the  length  of  the  hy- 
potenuse and  one  leg  of  a 
right  triangle  is  known, 
the  other  side  may  be 
found  by  squaring  the 
hypotenuse  and  squaring 
the  leg,  and  extracting  the 
square  root  of  their  dif- 
ference. 


Example.  —  If  the  hypotenuse  of  a  right  angle  triangle  is 
30"  and  the  base  is  18",  what  is  the  altitude? 


J 


302  =  30  X  30  =  900 

182  =  18  X  18  =  324 

900  _  324  =  576 

\/670=24".     Ana. 


Areas  of  Triangles 

The  area  of  a  triangle  may  be  found  when  the  length  of  the 
thi'ee  sides  is  given  by  adding  the  three  sides  together,  divid- 
ing by  2,  and  subtracting  from  this  sum  each  side  separately. 
Multiply  the  four  results  together  and  find  the  square  root  of 
their  product. 


68  VOCATIONAL    MATHEMATICS 

Example.  —  What  is   the  area  of   a  triangle  whose   sides 
measure  15,  16,  and  17  inches,  respectively  ? 

16 
16 


17  V24  X  9  X 8x7  =  \/l2096 

2^  V12096  =  109.98  sq.  in.    Ans. 

24  -  15  =  9 
24  -  16  =  8 
24-17  =  7 

Area  of  a  Triangle  =  ^  Base  x  Altitude 

Example.  —  What  is  the  area  of  a  triangle  whose  base  is 
17"  and  altitude  10"? 

1  ^ 

^  =  -  X  17  X  ;^  =  85  sq.  in.    Ans. 


EXAMPLES 

1.  A  ladder  17  ft.  long  standing  on  level  ground  reached  to 
a  window  12  ft.  from  the  ground.  If  it  is  assumed  that  the 
wall  is  perpendicular,  how  far  is  the  foot  of  the  ladder  from 
the  base  of  the  wall  ? 

2.  Find  the  area  of  a  triangular  sheet  of  metal  having  the 
base  81"  and  the  height  measured  from  the  opposite  angle  56". 

3.  Find  the  length  of  the  hypotenuse  of  a  right  triangle 
with  equal  legs  and  having  an  area  of  280  sq.  in. 

4.  Find  the  length  of  a  side  of  a  right  triangle  with  equal 
legs  and  an  area  of  72  sq.  in. 

5.  Find  the  hypotenuse  of  a  right  triangle  with  a  base  of 
8"  and  the  altitude  of  7". 

6.  What  is  the  area  of  a  triangle  whose  sides  measure  12, 
19,  and  21  inches  ? 

7.  What  is  the  altitude  of  an  isosceles  triangle  having  sides 
8  ft.  long  and  a  base  6  ft.  long  ? 


MENSURATION  69 

Quadrilaterals 

Four-sided  plane  figures  are  called  quadrilaterals.  Among 
them  are  the  trapezoid,  trajyezium,  rectangle,  rhotubus,  atid  rhom- 
boid. 


SquA&B  Rkctamolb  Rhomboid  Rhombus 


Trafbzium  Trapezoid  Parallelugbam 

Kinds  of  Quadrilaterals 

A  rectangle  is  a  quadrilateral  which  has  its  opposite  sides 
parallel  and  its  angles  right  angles.  Its  area  equals  the  prod- 
uct of  its  base  and  altitude. 

A=  ha 

A  trapezoid  is  a  quadrilateral  having  only  two  sides  parallel. 
Its  area  is  equal  to  the  product  of  the  altitude  by  one  half  the 
sum  of  the  bases. 

A=(b-\-c)xl^a  ^ 

In  this  formula    c  =  length  of  longest  side  J      ^       V 

b  =  length  of  shortest  side  /         i        \ 
a  =  altitude  * 

A  trapezium  is  a  four-sided  figure  with  no  two  sides  parallel. 
The  area  of  a  trapezium  is  found  by  dividing  the  trapezium 
into  triangles  by  means  of  a  diagonal.  Then  the  area  may*  be 
found  if  the  diagonal  and  perpendicular  heights  of  the  triangles 
are  known. 


70 


VOCATIONAL    MATHEMATICS 


Example.  —  In  the  trapezium  ABCD  if  the  diagonal  is  43' 
and  the  perpendiculars  11'  and  17',  respectively,  what  is  the 
area  of  the  trapezium  ? 


43  X  V- 
43  X 


H^ 


¥-  =  ^f 


:  236i-  sq.  ft.,  area  of  ABC 
:36ojsq.  ft.,areaof  ^Z>(7 
602    sq.  ft.,  total  area 
Ans. 


To  find   the   areas   of  irregular  figures, 
draw  the  longest  diagonal  and  upon  this 
diagonal  drop  perpendiculars  from  the  ver- 
tices of  the  figure.     These  perpendiculars  will  form  trapezoids 
and  right  triangles  whose  areas  may  be  determined  by  the  pre- 
ceding rules.     The  sum  of  the  areas  of  the  separate  figures  will 
give  the  area  of  the  whole  irregular  figure. 


Polygons 

A  plane  figure  bounded  by  straight  lines  is  a  polygon.  A 
polygon  which  has  equal  sides  and  equal  angles  is  a  regular 
polygon. 

The  apothem  of  a  regular  polygon  is  the  line  drawn  from  the 
center  of  the  polygon  perpen- 
dicular to  one  of  the  sides. 

A    five-sided    polygon    is    a 
pentagon. 

A    six-sided    polygon    is    a 
hexagon. 

An  eight-sided  polygon  is  an  octagon. 

The  shortest  distance  between  the  flats  of  a  regular  hexagon 
is  the  perpendicular  distance  between  two  opposite  sides,  and  is 
equal  to  the  diameter  of  the  inscribed  circle.  The  diameter  of 
the  -circumscribed  circle  is  the  long  diameter  of  a  regular  hexa- 
gon. 

The  perimeter  of  a  polygon  is  the  sum  of  its  sides. 


Pentagon 


Hexagon 


MENSURATION  71 

The  area  of  a  regular  polygon  equals  one  half  the  product  of 
the  apothem  and  the  perimeter. 

Formula  .1  =  ^(1/' 

In  this  formula  P  =  perimeter 

a  =  apothem 

Ellipse 

Only  the  approximate  circumference  of  an  ellipse  can  be  ob- 
tained. 

The  circumference  of  an  ellipse  equals  one  half  the  product  of 
the  sum  of  two  diameters  and  tt. 

If  c/i  =  major  diameter 

c/j  =  minor  diameter 
C  =  circumference 

then  c  =  4±^7r 

The  area  of  an  ellipse  is  equal  to  one  fourth  the  product  of 
the  major  and  minor  diameters  by  tt. 

If  A  =  area 

dj  =  major  diameter 
dj  =  minor  diameter 

then  A  =  TT^ 

4 

EXAMPLES 

1.  Find  the  area  of  a  trapezium  if  the  diagonal  is  93'  and 
the  perpendiculars  are  19'  and  33'. 

2.  What  is  the  area  of  a  trapezoid  whose  parallel  sides  are 
18  ft.  and  12  ft.,  and  the  altitude  8  ft.  ? 

3.  What  is  the  distance  around  an  ellipse  whose  major 
diameter  is  14"  and  minor  diameter  8"  ? 


72  VOCATIONAL    MATHEMATICS 

4.  In  the  map  of  a  country  a  district  is  found  to  have  two 
of  its  boundaries  approximately  parallel  and  equal  to  276  and 
216  miles.     If  the  breadth  is  100  miles,  what  is  its  area  ? 

5.  If  the  greater  and  lesser  diameters  of  an  elliptical  man- 
hole door  are  2'  9"  and  2'  6",  what  is  its  area  ? 

6.  Find  the  area  of  a  trapezium  if  the  diagonal  is  78"  and 
the  perpendiculars  18"  and  27". 

7.  The  greater  diameter  of  an  elliptical  funnel  is  4  ft.  6  in., 
and  the  lesser  diameter  is  4  ft.  (a)  What  is  its  area? 
(b)  How  many  square  feet  of  iron  will  it  contain  if  its  height 
is  16  ft.,  allowing  4"  for  the  seams  ? 

8.  What  is  the  area  of  a  pentagon,  whose  apothem  is  4i" 
and  whose  side  is  5"  ? 

Volumes 

The  volume  of  a  rectangular-shaped  bar  is  found  by  multi- 
plying the  area  of  the  base  by  the  length.  If  the  area  is  in 
square  inches,  the  length  must  be  in  inches. 

The  volume  of  a  cube  is  equal  to  the  cube  of  an  edge. 

The  contents  or  volume  of  a  cyliJidrical  solid  is  equal  to  the 
product  of  the  area  of  the  base  by  the  height. 

If  S  =  contents  or  capacity  of  cylinder 

R  =  radius  of  base 
H=  height  of  cylinder 
TT  =  3.1416+  or  ^/  (approx.) 
S  =  ttB^H 

Example.  —  Find  the  contents  of  a  cylindrical  tank  whose 
inside  diameter  is  14"  and  height  6'. 

H=6'  =  72" 

/S'  =  V  X  7  X  7  X  72  =  11,088  cu.  in. 


MENSURATION 


73 


The  Pyramid 

The  volume  of  a  puramid  eciuals  one 
third  of  the  product  of  the  area  of  the  base 
and  the  altitude. 

V=\ba 

The  volume  of  a  frustum  of  a  pyramid 
equals  the  product  of  one  third  the  alti- 
tude and  the  sum  of  the  two  bases  and  the 
square  root  of  the  product  of  the  bases. 

The  surface  of  a  regular  pyramid  is  equal  to  the  product  of 
the  perimeter  of  the  bases  and  one  half  the  slant  height. 

S=Px^sh 

The  Cone 

A  cone  is  a  solid  generated  by  a  right  triangle  revolving  on 
one  of  its  legs  as  an  axis. 

The  altitude  of  the  cone  is  the  perpendicular  distance  from 
the  base  to  the  apex. 

The  volume  of  a  cone  equals  the  product  of  the  area  of  the 
base  and  one  third  of  the  altitude. 

or  F=  .2618  D'H 

Example.  —  What  is  the  volume  of  a  cone  1^"    ^ 
in  diameter  and  4"  high  ? 

Area  of  base  =  .7864  x  | 


7.0686 


=  1.7671  sq.  in. 


F=  ^6181)2^ 

=  .2618  X  I  X  4  =  2.3662  cu.  In.   Ans, 

The  lateral  surface  of  a  cone  equals  one  half  the  product  of 
the  perimeter  of  the  base  by  the  slant  height. 


74 


VOCATIONAL    MATHEMATICS 


Example.  —  What  is  the  surface  of  a  cone  having  a  slant 
height  of  36  in.,  and  a  diameter  of  14  in.  ? 


3^ 


14  X  V  =  44" 


^i-^^  =  702  sq.  in.    Ans. 


Frustum  of  a  Cone 

The  frustum  of  a  cone  is  the  part  of  a  cone  included  between 
the  base  and  a  plane  or  upper  base  which  is  parallel  to  the 
lower  base. 

The  volume  of  a  frustum  of  a  cone  equals  the  product  of  one 
third  of  the  altitude  and  the  sum  of  the  two  bases  and  the 
square  root  of  their  product. 

When  H  =  altitude 

B^  =  upper  base 
B  =  lower  base 
V=iH{B-hB'-\-  VBB') 

The  lateral  surface  of  a  frustum  of  a  cone  equals  one  half  the 
product  of  the  slant  height  and  the  sum  of  the  perimeters 
of  the  bases. 

The  Sphere 

The  volume  of  a  sphere  is  equal  to 

3 
where  B  is  the  radius. 

The  surface  of  a  sphere  is  equal  to 

The  Barrel 

To  find  the  cubical  contents  of  a  barrel,  (1)  multiply  the 
square  of  the  largest  diameter  by  2,  (2)  add  to  this  product 


MENSURATION  75 

the  square  of  the  head  diameter,  and  (3)  multiply  this  sum  by 
the  length  of  the  barrel  and  that  product  by  .2618. 

Example.  —  Find  the  cubical  contents  of  a  barrel  whose 
largest  diameter  is  21"  and  head  diameter  18",  and  whose 
length  is  33". 

V=  1{D'^  X  2)  +  d'^]  X  L  X  .2618 
31)798 
.2618 


21*  = 

=  441  X 

2  =  882 

182  = 

=  324 

.324 

1206 

33 

3618 

3618 

10419.11  cu.  in 
10419.11 


39798  281 


=  46.10  gal.    Ans. 


Similar  Figures 

Similar  figures  are  figures  that  have  exactly  the  same  shape. 
The  areas  of  similar  figures  have   the   same    ratio   as  the 
squares  of  their  corresponding  dimensions. 

Example.  —  If  two  boilers  are  15'  and  20'  in  length,  what  is 
the  ratio  of  their  surfaces  ? 

J^  =  I,  ratio  of  lengths 

—  =  — ,  ratio  of  surfaces 
42      16 

One  boiler  is  ^5  as  large  as  the  other.    Ans. 

The  volumes  of  similar  figures  are  to  each  other  as  the  cubes 
of  their  corresponding  dimensions. 

Example.  —  If   two  iron  balls  have  8"  and  12"  diameters, 
respectively,  what  is  the  ratio  of  their  volumes  ? 

^  =  f ,  ratio  of  diameters 

=  /y,  ratio  of  their  volumes.     Am. 

One  ball  weighs  ^  as  much  as  the  other. 


76  VOCATIONAL    MATHEMATICS 

EXAMPLES 

1.  Find  the  volume  of  a  rectangular  iron  bar  8''  by  10" 
and  4'  long. 

2.  Find  the  weight  of  a  rectangular  steel  bar  31"  x  49"  x  3" 
thick,  if  the  metal  weighs  .28  lb.  per  cubic  inch. 

3.  The  radius  of  the  small  end  of  a  bucket  is  4  in.  Water 
stands  in  the  bucket  to  a  depth  of  9  in.,  and  the  radius  of  the 
surface  of  the  water  is  6  in.  (a)  Find  the  volume  of  the 
water  in  cubic  inches.  (6)  Find  the  volume  of  the  water  in 
gallons  if  a  cubic  foot  contains  7.48  gal. 

4.  What  is  the  volume  of  a  steel  cone  2i"  in  diameter  and 
6"  high  ? 

5.  Find  the  contents  of  a  barrel  whose  largest  diameter  is 
22",  head  diameter  18",  and  height  35". 

6.  What  is  the  volume  of  a  sphere  8"  in  diameter  ? 

7.  What  is  the  volume  of  a  pyramid  with  a  square  base, 
4"  on  a  side  and  11"  high  ? 

8.  What  is  the  surface  of  a  steel  cone  with  a  6"  diameter 
and  14"  slant  height  ? 

9.  Find  the  surface  of  a  pyramid  with  a  perimeter  of  18" 
and  a  slant  height  of  11". 

10.  Find  the  volume  of  a  cask  whose  height  is  3^'  and  the 
greatest  radius  16"  and  the  least  radius  12",  respectively. 

11.  What  is  the  weight  of  a  cast-iron  cylinder  2.75"  in 
diameter  and  12j"  long,  if  cast  iron  weighs  450  lb.  per 
cu.  ft.  ? 

12.  How  many  gallons  of  water  will  a  round  tank  hold 
which  is  4  ft.  in  diameter  at  the  top,  5  ft.  in  diameter  at  the 
bottom,  and  8  ft.  deep  ?     (231  cu.  in.  =  1  gal.) 

13.  What  is  the  volume  of  a  cylindrical  ring  having  an 
outside  diameter  of  6 J",  an  inside  diameter  of  5^",  and  a 
height  of  3|"  ?    What  is  its  outside  area  ? 


MENSURATION  77 

14.  A  sphere  has  a  circumference  of  8.2467".  (a)  What  is 
its  area?     (6)  What  is  its  volume ? 

15.  If  it  is  desired  to  make  a  conical  oil  can  with  a  base 
3.5"  in  diameter  to  contain  \  pint,  what  must  the  approximate 
height  be  ? 

16.  What  is  the  area  of  one  side  of  a  flat  ring  if  the  inside 
diameter  is  2^"  and  the  outside  diameter  4J"  ? 

17.  There  are  two  balls  of  the  same  material  with  diameters 
4"  and  1",  respectively.  If  the  smaller  one  weighs  3  lb.,  how 
much  does  the  larger  one  weigh  ? 

18.  If  the  inside  diameter  of  a  ring  must  be  5  in.,  what 
must  the  outside  diameter  be  if  the  area  of  the  ring  is  6.9 
sq.  in.  ? 

19.  How  much  less  paint  will  it  take  to  paint  a  wooden 
ball  4"  in  diameter  than  one  10"  in  diameter  ? 

20.  What  is  the  weight  of  a  brass  ball  3J"  in  diameter  if 
brass  weighs  .303  lb.  per  cubic  inch  ? 

21.  A  cube  is  19"  on  its  edge,  (a)  Find  its  total  area. 
(6)  Its  volume. 

22.  If  the  area  of  a  J"  pipe  is  .049  sq.  in.,  what  will  be  the 
diameter  of  a  pipe  having  8  times  the  area? 

23.  What  is  the  weight  of  a  cast-iron  cylinder  2.75'  in 
diameter  and  12}'  long,  if  the  cast  iron  weighs  450  lb.  per 
cubic  foot  ? 

24.  A  conical  funnel  has  an  inside  diameter  of  19.25"  at  the 
base  and  is  43"  high  inside,  (a)  Find  its  total  area.  (6)  Find 
its  cubical  contents. 

25.  If  a  bar  3"  in  diameter  weighs  24.03  lb.  per  foot  of 
length,  what  must  be  the  weight  per  foot  of  a  bar  3"  square  of 
the  same  material? 


CHAPTER   III 

Reading  a  Blue  Print 

Every  skilled  worker  in  wood  or  metal  must  know  how  to 
read  a  "  blue  print,"  which  is  the  name  given  to  working  plans 


Screw  Table  Frame 

Simple  Blue  Prints  or  Working  Drawings 

and  drawings  with  white  lines  upon  a  blue  background.     The 
blue   print   is   the  language  which  the  architect  uses  to   the 

78 


WORKING   DRAWINGS  79 

builder,  the  machinist  to  the  pattern  maker,  the  engineer  to  the 
foreman  of  construction,  and  the  designer  to  the  workman, 
Through  following  the  directions  of  the  blue  print  the  carpenter, 
metal  worker,  and  mechanic  are  able  to  produce  the  object 
wanted  by  the  employer  and  his  designer  or  draftsman. 

Two  views  are  usually  necessary  in  every  working  drawing, 
one  the  plan  or  top  view  obtained  by  looking  down  upon  the 
object,  and  the  other  the  elevation  or  front  view.  When  an  object 
is  very  complicated,  a  third  view,  called  an  end  or  profile  view, 
is  shown. 

All  the  information,  such  as  dimensions,  etc.,  necessary  to  construct 
whatever  is  represented  by  the  blue  print,  must  be  supplied  on  the  draw- 
ing. If  the  blue  print  represents  a  machine  it  is  necessary  to  show  all  the 
parts  of  the  machine  put  together  in  their  proper  places.  This  is  called 
an  assembly  drawing.  Then  there  must  be  a  drawing  for  each  part  of 
the  machine,  giving  information  as  to  the  size,  shape,  and  number  of 
the  pieces.  Then  if  there  are  interior  sections,  these  must  be  represented 
in  section  drawings. 

Drawing  to  Scale 

As  it  is  impossible  to  draw  most  objects  full  size  on  paper,  it 
is  necessary  to  make  the  drawings  proportionately  smaller. 
This  is  done  by  making  all  the  dimensions  of  the  drawing  a 
certain  fraction  of  the  true  dimensions  of  the  object.  A  draw- 
ing made  in  this  way  is  called  drawn  to  scale. 


Triangular  Scale 

The  dimensions  on  the  drawing  are  designated  the  actual 
size  of  the  object  —  not  of  the  drawing.  If  a  drawing  were 
made  of  an  iron  bolt  25  inches  long,  it  would  be  inconvenient 
to  represent  the  actual  size  of  the  bolt,  and  the  drawing  might 
be  made  half  or  ipiarter  the  size  of  the  bolt,  but  the  length 
would  read  on  the  drawing  25  inches. 


80  VOCATIONAL    MATHEMATICS 

In  making  a  drawing  "to  scale,"  it  becomes  very  tedious  to 
be  obliged  to  calculate  all  the  small  dimensions.  In  order  to 
obviate  this  work  a  triangular  scale  is  used.  It  is  a  ruler  with 
the  different  scales  marked  on  it.  By  practice  the  student  will 
be  able  to  use  the  scale  with  as  much  ease  as  the  ordinary 
ruler. 

QUESTIONS   AND   EXAMPLES 

1.  Tell  what  is  the  scale  and  the  length  of  the  drawing  of 
each  of  the  following : 

a.  An  object  14"  long  drawn  half  size. 

h.  An  object  26"  long  drawn  quarter  size. 

c.  An  object  34"  long  drawn  one  third  size. 

d.  An  object  41"  long  drawn  one  twelfth  size. 

2.  If  a  drawing  made  to  the  scale  of  |"  =  1  ft.  is  reduced  \  in 
size,  what  will  the  new  scale  be  ? 

3.  The  scale  of  a  drawing  is  made  \  size.  If  it  is  doubled, 
how  many  inches  to  the  foot  will  the  new  scale  be  ? 

4.  On  the  yy  scale,  how  many  feet  are  there  in  ^d> 
inches  ? 

5.  On  the  y  scale,  how  many  feet  are  there  in  26  inches  ? 

6.  On  the  ^"  scale,  how  many  feet  are  there  in  21  inches  ? 

7.  If  the  drawing  of  a  bolt  is  made  -J-  size  and  the  length  of 
the  drawing  is  8^',  what  will  it  measure  if  made  to  scale  3" 
=  lft.? 

8.  What  will  be  the  dimensions  of  the  drawing  of  a  machine 
shop  582'  by  195'  if  it  is  made  to  a  scale  of  yV  =  1  ft.  ? 

Arithmetic  and  Blue  Prints.  —  Mechanics  are  obliged  to  read  from  blue 
prints.  In  order  to  verify  the  necessary  dimensions  in  the  detail  of  the 
work,  it  becomes  necessary  to  do  more  or  less  addition  and  subtraction  of 
the  given  dimensions.  This  involves  ability  to  add, and  subtract  mixed 
numbers  and  fractions. 


METHODS  OF  SOLVING   EXAMPLES  81 

Methods  of  Solving  Examples 

Every  mechanical  problem  or  operation  has  two  distinct 
sides :  the  collecting  of  data  and  the  solving  of  the  problem. 

The  first  part,  the  collecting  of  data,  demands  a  knowledge 
of  the  materials  and  conditions  under  which  the  problem  is 
given,  and  calls  for  considerable  judgment  as  to  the  necessary 
accurateness  of  the  work. 

There  are  three  ways  by  which  a  problem  may  be  solved : 

1.  Exact  method. 

2.  Rule  of  thumb  method,  by  the  use  of  a  two-foot  rule  or 
a  slide  scale. 

3.  By  means  of  tables. 

The  exact  method  of  solving  a  problem  in  arithmetic  is  the 
one  usually  taught  in  school  and  is  the  method  obtained  by 
analysis.  Every  one  should  be  able  to  solve  a  problem  by  the 
exact  method. 

The  rule  of  thumb  method. — Many  of  the  problems  that  arise 
in  industrial  life  have  been  met  before  and  very  careful  judg- 
ment has  been  exercised  in  solving  them.  As  the  result  of 
this  experience  and  the  tendency  to  abbreviate  and  devise 
shorter  methods  that  give  sufficiently  accurate  results,  we  find 
many  rule  of  thumb  methods  used  by  the  mechanics  in  daily 
life.  The  exact  method  would  involve  considerable  time  and 
the  use  of  pencil  and  paper,  whereas  in  cases  that  are  not  too 
complicateid  the  two-foot  rule  or  the  slide  scale  will  give  a 
quick  and  accurate  result. 

In  solving  problems  involving  the  addition  and  subtraction  of  fractions, 
by  the  rule  of  thumb,  use  the  two-foot  rule  or  steel  scale  to  carry  on  the 
computation.  To  illustrate  :  if  we  desire  to  add  \  and  ^,  place  the 
thumb  on  \  division,  then  slide  (move)  the  thumb  along  a  division  cor- 
responding to  ^,  and  then  read  the  number  of  divisions  passed  over  by 
the  thumb.  In  this  case  the  result  is  ^.  For  fractions  involving  ^,  ^^, 
^,  and  j^  use  the  steel  scale.  The  majority  of  machinists,  carpenters, 
etc.,  use  this  method  of  sliding  the  thumb  over  the  rule,  in  adding  and 
subtracting  inches  and  fractions  of  inches. 


82  VOCATIONAL  MATHEMATICS 

The  use  of  tables.  —  In  the  commercial  and  industrial  world 
the  tendency  is  to  do  a  thing  in  the  quickest  and  the  most 
economical  way.  To  illustrate  :  hand  labor  is  more  costly 
than  machine  work,  so  wherever  possible,  machine  work  is 
substituted  for  hand  labor.  The  same  condition  applies  to 
the  calculations  that  are  used  in  the  shop.  The  methods  of 
performing  calculations  are  the  most  economical  —  that  is, 
the  quickest  and  most  accurate  —  that  the  ordinary  mechanic 
is  able  to  perform.  Since  a  great  many  of  the  problems  in 
calculation  that  arise  in  the  daily  experience  of  the  mechanic 
are  about  standardized  pieces  of  metal  and  repeat  themselves 
often,  it  is  not  necessary  to  wot-k  them  each  time  if  results  are 
kept  on  file  when  they  are  once  solved.  This  tiling  is  done 
by  means  of  tables  that  are  made  from  these  problems. 

See  pages  105,  117,  and  121,  for  tables  used  in  this  book. 


PART  II  — MATHEMATICS    FOR    CARPENTERING 
AND  BUILDING 

CHAPTER   IV 
MEASURING  LUMBER 

The  carpenter  or  builder  is  often  required  to  give  an  estimate 
of  the  cost  of  the  work  to  be  done  for  his  prospe(;tive  custom- 
ers. People  who  contemplate  building  have  several  estimates 
submitted  to  them  by  different  builders  and  generally  give  the 
work  to  the  lowest  bidder.  An  architect  usually  draws  plans 
of  the  building  and  from  these  plans  the  contractor  or  builder 
makes  his  estii^ate  of  the  cost.  In  doing  repairing  or  cabinet- 
making  the  carpenter  makes  his  own  plans  and  estimates.  In 
order  to  make  a  proper  estimate  of  the  cost,  one  must  know  the 
market  price  of  materials,  the  cost  of  labor,  the  amount  of 
material  needed,  and  the  length  of  time  required  to  do  the  work. 

Preparation  of  Wood  for  Building  Purposes 

In  winter  the  forest  trees  are  cut  and  in  the  spring  the  logs  floated 
down  the  rivers  to  sawmills,  where  they  are  sawed  into  boards  of  different 
thickness.  To  square  the  log,  four  slabs  are  first  sawed  off.  After  these 
slabs  are  off,  the  remainder  is  sawed  into  boards. 

As  soon  as  the  boards  or  planks  are  sawed  from  the  logs,  they  are  piled 
on  prepared  foundations  in  the  open  air  to  season.  Each  layer  is  separated 
from  the  one  above  by  a  crosspiece,  called  a  strap,  in  order  to  allow  free  cir- 
culation of  air  about  each  board  to  dry  it  quickly  and  evenly.  If  lumber 
were  to  be  piled  up  without  the  strips,  one  board  upon  another,  the  ends 
of  the  pile  would  dry  and  the  center  would  rot.  This  seasoning  or  drying 
out  of  the  sap  usually  lasts  several  months.  *'  Air  dried  "  lumber  is  used 
for  most  building  purposes  except  in  buildings  or  places  where  there  is 
a  warm,  dry  atmosphere. 

88 


84  VOCATIONAL  MATHEMATICS 

Wood  that  is  to  be  subject  to  a  warm  atmosphere  has  to  be  artificially 
dried.  This  artificially  dried  or  kiln-dried  lumber  has  to  be  dried  to  a  point 
in  excess  of  that  of  the  atmosphere  in  which  it  is  to  be  placed  after  being 
removed  from  the  kiln.  This  process  of  drying  must  be  done  gradually 
and  evenly  or  the  boards  may  warp  and  then  be  unmarketable. 

Definitions 

Board  Measure.  —  A  board  one  inch  or  less  in  thickness  is  said 
to  have  as  many  board  feet  as  there  are  square  feet  in  its  surface. 
If  it  is  more  than  one  inch  thick,  the  number  of  board  feet  is 
found  by  multiplying  the  number  of  square  feet  in  its  surface 
by  its  thickness  measured  in  inches  and  fractions  of  an  inch. 

The  number  of  board  feet  =  length  {in  feet)  x  width  (in  feet)  x  thick- 
ness (in  inches). 

Board  measure  is  used  for  plank  measure.  A  plank  2"  thick,  10"  wide, 
and  15'  long,  contains  twice  as  many  square  feet  (board  measure)  as 
a  board  1"  thick  of  the  same  width  and  length. 

To  measure  a  board  that  tapers,  the  width  is  taken  at  the 
middle,  where  it  is  one  half  the  sum  of  the  widths  of  the 
ends. 

Boards  are  sold  at  a  certain  price  per  hundred  (C)  or  per 
thousand  (M)  board  feet. 

The  term  lumber  is  applied  to  pieces  not  more  than  four 
inches  thick  ;  timber  to  pieces  more  than  four  inches  thick  ;  but 
a  large  amount  taken  together  often  goes  by  the  general  name 
of  lumber.  A  piece  of  lumber  less  than  an  inch  and  a  half 
thick  is  called  a  board  and  a  piece  from  one  inch  and  a  half  to 
four  inches  thick  a  plank. 

Rough  Stock  is  lumber  the  surface  of  which  has  not  been 
dressed  or  planed. 

The  standard  lengths  of  pieces  of  lumber  are  10,  12,  14,  16, 
18  feet,  etc. 

In  measuring  and  marking  large  lots  of  lumber  in  which 
there  are  a  number  of  pieces  containing  a  fraction  of  a  foot,  in  case 
of  one  half  foot  or  more,  1  is  added,  and  in  case  of  less  than  one 
half  foot,  it  is  disregarded.     This  is  especially  true  with  boards. 


CARPENTERING  AND  BUILDING  85 

A  board  1"  x  4"  x  16'  contains  6 J  feet  (board  measure).  Two 
boards  would  be  marked  6  ft.,  and  every  third  board  would  be  marked 
(i  ft.  So  two  boards  may  be  exactly  the  same  length  and  one  marked  6  ft. 
and  the  other  0  ft.  In  the  purchasing  of  a  single  board  there  might  be  a 
small  undercharge  or  overcliarge,  but  in  large  lots  the  average  would  be 
struck. 

EXAMPLES 

1.  How  many  board  feet  in  a  board  1  in.  thick,  15  in.  wide, 
and  15  ft.  long  ? 

2.  How  many  board  feet  of  2-inch  planking  will  it  take  to 
make  a  walk  3  feet  wide  and  4  feet  long  ? 

3.  A  plank  19'  long,  3"  thick,  10"  wide  at  one  end  and  12" 
wide  at  the  other,  contains  how  many  board  feet  ? 

4.  Find  the  cost  of  7  2-inch  planks  12  ft. long,  16  in.  wide 
at  one  end,  and  12  in.  at  the  other,  at  S  0.08  a  board  foot. 

5.  At  $  12  per  M,  what  will  be  the  cost  of  2-inch  plank  for 
a  3  ft.  6  in.  sidewalk  on  the  street  sides  of  a  rectangular  corner 
lot  56  ft.  by  106  ft.  6  in.  ? 

Quick  Method  for  Measuring  Boards 

To  measure  boards  1"  thick,  multiply  length  in  feet  by 
width  in  inches  and  divide  by  12,  and  the  result  will  be  the 
board  measure  in  feet. 

For  boards  IJ"  thick,  add  one  quarter  of  the  quotient  to  the 
result  as  above. 

For  boards  1|"  thick,  add  one  half  of  the  quotient  to  the 
r(!sult  as  found  above. 

For  plank  2"  thick,  divide  by  6  instead  of  12. 

For  plank  3"  thick,  divide  by  4  instead  of  12. 

For  plank  4"  thick,  divide  by  3  instead  of  12. 

For  timber  6"  thick,  divide  by  2  instead  of  12. 

EXAMPLES 

1.  Find  the  number  of  board  feet  in  10  planks,  3"  thick, 
12"  wide,  14'  long. 


86 


VOCATIONAL  MATHEMATICS 


2.  Find  the  number  of  board  feet  in  4  timbers,  8"  thick, 
10"  wide,  17'  long. 

3.  Find  the  number  of  board  feet  in  18  joists,  2"  thick, 
4"  wide,  14'  long. 

4.  Find  the  number  of  board  feet  in  16  beams,  10"  thick, 
12"  wide,  11'  long. 

5.  Find  the  number  of  board  feet  in  112  boards,  J"  thick, 
.8"  wide,  14'  long. 

Note.  — Ordinarily  fractions  of  a  foot  less  than  one  half  are  omitted, 
but  when  the  fraction  is  one  half  or  larger  it  is  reckoned  as  a  foot.  This 
is  sufficiently  accurate  for  all  practical  purposes. 

Board  measure  of  one  lineal  foot  of  timber  may  be  found 
from  the  following  table  : 

Contents  (Board  Measure)  of  One  Lineal  Foot  of  Timber 


5r. 

T 

HICKNE88    IN    InOIIES 

2 

3 

4 

5 

6 

7 

8 

9 

10 

11 

12 

13 

14 

18 

3. 

4.5 

6. 

7.5 

9. 

10.5 

12. 

13.5 

15. 

16.5 

18 

19.5 

21. 

17 

2.83 

4.25 

5.66 

7.08 

8.5 

9.92 

11.33 

12.75 

14.17 

15.58 

17 

18.42 

19.83 

16 

2.67 

4. 

5.33 

6.67 

8. 

9.33 

10.67 

12. 

13.33 

14.67 

16 

17.3 

18.66 

15 

2.5 

3.75 

5. 

6.25 

7.5 

8.75 

10. 

11.25 

12.5 

13.75 

15 

16.25 

17.6 

14 

2.33 

3.5 

4.67 

5.83 

1 . 

8.17 

9.33 

10.5 

11.67 

12.83 

14 

15.17 

16. ()6 

13 

2.17 

3.25 

4.33 

5.42 

6.5 

7.58 

8.67 

9.75 

10.83 

11.92 

13 

14.08 

12 

2. 

3. 

4. 

5. 

6. 

7. 

8. 

9. 

10. 

11. 

12 

11 

1.83 

2.75 

3.67 

4.58 

5.5 

6.42 

7.33 

8.25 

9.17 

10.08 

10 

1.67 

2.5 

3.33 

4.17 

5. 

5.88 

6.67 

7.5 

8.33 

9 

1.5 

2.25 

3. 

3.75 

4.5 

5.25 

6. 

6.75 

8 

1.33 

2. 

2.67 

3.33 

4. 

4.67 

5.33 

7 

1.17 

1.75 

2.33 

2.92 

3.5 

4.08 

6 

1. 

1.5 

2. 

2.5 

3. 

5 

.83 

1.25 

1.67 

2.08 

4 

.67 

1. 

1.33 

3 

.5 

.75 

2 

.33 

CARPENTERING  AND  BUILDING 


87 


To  ascertain  the  contents  of  a  piece  of  timber,  find  in  the 
table  the  contents  of  one  foot  and  multiply  by  the  length  in 
feet  of  the  piece. 

EXAMPLES 
By  means  of  the  above  table  find  the  board  measure  of  the 

following : 

1.  One  timber  8"  x  9",  14'  long. 

2.  One  timber  8"  X  11",  13'  long. 

3.  One  timber  9"  x  10",  11'  long. 

4.  One  timber  8"  x  10",  16'  long. 

5.  One  timber  7"  x  9",  11'  long. 

6.  Two  planks  2"  x  3",  10'  long. 

7.  Two  planks  4"  x  4",  13'  long. 

8.  Two  timbers  10"  x  11",  15'  long. 

9.  Two  timbers  10"  x  18",  14'  long. 
10.  Two  timbers  8"  X  16",  13'  long. 

The  weight  per  cubic  foot  of  different  woods  can  be  readily 
seen  from  the  following  table : 


Weight  of  One  Cubic  Foot  of  Timber 

Wood 

Weight 
I'ER  CiT.  Ft. 

Wood 

Wbiuut 
PER  Cv.  Ft. 

White  Pine  .     . 
Georgia  Pine 
Hemlock  .     .     . 
Cypress    .     .     . 
Spruce      .     .     . 
White  Oak   .     . 
Red  Oak  .     .     . 
Maple.     .     .     . 

28  1b. 
38  1b. 
24  1b. 
331b. 
28  1b. 
481b. 
46  1b. 
42  lb. 

Whitewood    .... 

Ash 

Hickory    .     •     .     .     . 

Chestnut 

Cedar 

Birch   ....... 

Ebony 

Boxwood 

30  lb. 
45  1b. 
48  1b. 
36  1b. 
39  1b. 
411b. 
76  1b. 
70  lb. 

Lumber  is  bought  and  sold  in  the  log  by  cubic  measure. 
Lumber  used  for  framing  buildings  and  for  building  bridges, 
docks,  and  ships  is  also  sold  by  the  cubic  foot. 


88  VOCATIONAL  MATHEMATICS 

The  rule  that  is  most  extensively  used  for  computing  the 
contents  in  board  feet  of  a  log  is  as  follows : 

Rule.  —  SvMract  4  inches  from  the  diameter  of  the  log  at  the 
small  end,  square  one  quarter  of  the  remainder,  and  multiply  the 
result  by  the  length  of  the  log  in  feet. 

EXAMPLES 
Find  the  board  feet  of  lumber  in  the  following  logs : 
Diameter  in  inches       ^-      2.      3.      4.      5.      6.      7.      8.      9. 
at  small  end:  12     13     14     15     16     17     18     20     24 

Length  in  feet :  10     12     14     16     18     20     16     20    24 


CHAPTER   Y 
CONSTRUCTION 

Excavations.  —  After  the  })lans  for  a  building  are  drawn  by 
the  architect  and  the  work  given  to  the  contractor  on  a  bid, 
usually  the  lowest,  the  work  of  excavating  begins.  In  esti- 
mating excavations  the  cubic  yard,  or  27  cubic  feet,  is  used. 

EXAMPLES 

1.  What  will  it  cost  to  excavate  a  cellar  that  is  32'  x  28' 
and  5'  deep  at  34  cents  per  cubic  yai-d  ? 

2.  What  will  the  cost  be  of  excavating  a  lot  112'  by  58'  and 
averaging  12'  deep  at  $1.65  per  yard  ? 

3.  In  excavating  a  tunnel  374,166  cubic  feet  of  earth  were 
removed.  If  the  length  of  the  tunnel  was  492  ft.  and  th« 
width  39  ft.,  what  was  the  height  of  the  tunnel  ? 

4.  How  many  cubic  yards  of  earth  must  be  removed  to 
build  a  cellar  for  a  house  when  the  measurements  inside  the 
wall  are  28'  long  and  16'  wide,  the  wall  being  1'.8"  thick 
and  8'  deep,  with  2'  of  the  wall  above  the  ground  level? 

5.  In  making  a  bid  on  some  excavating  a  contractor  notes 
that  the  excavation  is  in  the  shape  of  a  rectangle  8'  deep,  11' 
wide  at  the  top  and  bottom,  and  483'  long.  What  will  it  cost 
him  to  excavate  it  at  29  cents  per  cubic  yard  ?  What  must 
his  bid  be  to  make  10  %  profit  ? 

Frame  and  Roof 

After  the  excavation  is  finished  and  the  foundation  laid,  the 
construction  of  the  building  itself  is  begun.  On  the  top  of  the 
foundation  a  large  timber  called  a  sill  is  placed.     The  timbers 


90 


VOCATIONAL  MATHEMATICS 


running  at  right  angles  to  the  front  sill  are  called  side  sills ; 
The  sills  are  joined  at  the  corners  by  a  half-lap  joint  and  held 
together  by  spikes. 


a.  Outside  studding 

b.  Rafters 

c.  Plates 

il.  Ceiling  joists 


de.  Second  floor  joists 

def.  First  floor  joists 

g.  Girder  or  cross  sill 

h.  Sills 


i.   Sheathing 

j.   Partition  studs 

k.  Partition  heads 

l.   Piers  1 


Then  the  building  has  its  walls  framed  by  placing  corner 
posts  of  4"  by  6"  on  the  four  corners.  Between  these  corner 
posts  there  are  placed  smaller  timbers  called  studding,  2"  by 
4",  16"  apart.  Later  the  laths,  4'  long,  are  nailed  to  this  stud- 
ding. The  upright  timbers  are  mortised  into  the  sills  at  the 
bottom.  When  these  uprights  are  all  in  position  a  timber 
called  a  plate  is  placed  on  the  top  of  them  and  they  are  spiked 
together. 

On  the  top  of  the  plate  is  placed  the  roof.  The  principal 
timbers  of  the  roof  are  the  rafters.     Different  roofs  have  a  dif- 


1  If  made  of  brick  or  stone,  '' shones "  or  "supports  "  if  made  of  wood. 


CARPENTERING  AND  BUILDING 


91 


fereiit  pitcli  or  slope  —  that  is,  form 
different  angles  with  the  plate.  To 
get  the  desired  pitch  the  carpenter 
uses  the  steel  square. 

When  we  speak  of  the  pitch  of  a  roof  we 
mean  the  slope  or  slant  of  the  roof.  A 
roof  with  oi>€  half  pitch  means  that  the 
height  of  the  rid<j;e  of  the  roof  above  the 
level  of  the  plate  is  equal  to  one  half  of  the 

width  of  the  roof,  or  —  =  H. 
2 

If  the  width  of  a  building  is  1(5  feet  and       mo  <j  87054321 

the  roof  is  one  quarter  pitch,  it  means  that  the  height  of  the  ridge  of  the 

roof  above  the  plate  is  4  feet,  or  -    =  R. 

EXAMPLES 
Give  the  height  of  the  ridge  of  the  roof  above  the  level  of 
the  plate  of  the  following  buildings : 


Pitch 

Width  or  BriLDiNo 

Pitch 

Wi 

UTH   OK    BriLDINU 

1.    i 

32  feet 

5.   \ 

36  feet 

2    \ 

40  feet 

6-    1 

48  feet 

3-    i 

86  feet 

7-    1 

28  feet 

4.    1 

48  feet 

8.    A 

34  feet 

The  height  of  the  ridge  of  the 
roof  above  the  level  of  the  plate 
is  the  rise  of  rafter.  The  dis- 
tance from  where  the  rafter  in- 
tersects the  outer  edge  of  the 
plate  to  a  point  on  the  plate  di- 
rectly beneath  the  peak  is  called 
the  run  of  rafter. 

'I  he  figure  represents  a  frame,  A^ 
B.  C\  D,  with  rafters  E  and  JP,  braces 
(t  mid  H.  The  frame  is  12  feet  high 
nnd  12  feet  wide. 

Kafter  E  is  shown  as  a  half-pitch 
roof,  meaning  that  its  rise  is  equal  to 


92  VOCATIONAL  MATHEMATICS 


iiuwM»wMM^iimitMww^-W1<iH<trimWfHBffl>kn|wtiiHi'iiviPj<ti»m|imHi|WtHMHJ|'iia)tii^  half       t  ll  6 

"I  width  of  the 


building,  or 
Cakpenter's  Steel  Square  q  ^^^^ 

Rafter  F 
is  shown  as  a  third-pitch  roof  and  its  rise  is  equal  to  one  third  the 
width  of  the  building,  or  4  feet. 

The  proportion  between  the  roof  and  the  plate  always  holds 
good. 

To  find  the  length  of  a  rafter  for  a  half-pitcli  roof,  take 
lialf  of  the  width  of  the  building  (6')  or  3  to  1  on  the 
blade  of  the  square  and  rise  (6'),  or  1  to  3  on  the  tongue 
of  the  square.  The  distance  in  a  straight  line  from  6 
on  both  blade  and  tongue  is  8|,  which  is  the  length  of 
the  rafter  needed,  as  shown  from  2  to  2.  The  measure 
must  be  taken  on  the  line  indicated  by  the  arrow  I 
(see  diagram  on  page  91),  not  on  the  top  of  the  rafter.  It 
is  usual  to  cut  away  a  piece  2  inches  thick  from  the  part 
of  the  rafter  that  projects  past  the  side  walls,  —  the  part 
to  which  the  cornice  is  nailed,  or  2  inches  down  from  the 
top,  —  which  is  the  line  to  measure  for  the  length  of  the 
rafter. 

To  find  the  cut  (bevel)  of  the  top  and  bottom  of  the  rafter, 
lay  the  square  flat  on  the  side  of  the  rafter  with  the  blades 
toward  the  bottom  and  the  tongue  toward  the  top,  with  the 
figure  6  on  the  blade  and  the  figure  4  on  the  tongue  at  the 
edge  of  the  rafter.  Mark  along  the  blade  for  the  bottom  and 
cut  along  the  tongue  for  the  top  cut.  This  gives  the  bevel  for 
the  top  and  bottom  cuts  for  a  third-pitch  roof.  With  6  on 
both  blade  and  tongue  you  get  the  bevel  for  a  half-pitch  roof 
or  a  miter  cut  for  any  square  purpose. 

Mark  the  run  on  the  blade  and  the  rise  on  the  tongue  and  then 
measure  across  from  one  figure  to  the  other  and  you  have  the 
length  of  any  rafter  or  brace.  Take  the  brace  H  in  the  diagram 
on  page  91,  for  example.  It  has  4-foot  rise  and  2-foot  6-inch 
run.    Its  length  is  found  in  the  same  way.     By  measuring 


CARPENTERING  AND  BUILDING  93 

from  4  on  the  blade  to  2^  on  the.  tongue  you  find  4  feet  8^ 
inches  the  length  of  the  brace. 

In  measuring  a  brace  you  must  mark  the  length  on  the 
outer  edge  instead  of  2  inches  inside  (like  the  rafter),  for  the 
tenon  is  usually  made  to  extend  to  the  extreme  point.  The 
bevel  at  each  end  is  obtained  by  the  square  in  the  same  man- 
ner as  with  the  rafter. 

Lathing 

Laths  are  thin  pieces  of  wood,  4  ft.  long  and  IJ  in.  wide, 
upon  which  the  plastering  of  a  house  is  laid.  They  are  usu- 
ally put  up  in  bundles  of  one  hundred.  They  are  nailed  J  in. 
apart  and  fifty  will  cover  about  30  sq.  ft. 

EXAMPLES 

1.  At  30  cents  per  square  yard  what  will  it  cost  to  lath  and 
plaster  a  wall  12  ft.  by  15  ft.  ? 

2.  At  45  cents  per  square  yard  what  will  it  cost  to  lath  and 
plaster  a  wall  18  ft.  by  16  ft.  ? 

4 

3.  What  will  it  cost  to  lath  and  plaster  a  room  (including 
walls  and  ceiling)  16  ft.  square  by  12  ft.  high,  allowing  34  square 
feet  for  windows  and  doors,  at  40  cents  per  square  yard  ? 

4.  What  will  it  cost  to  lath  and  plaster  the  following  rooms 
at  42J  cents  per  square  yard  ? 

a.  16' X 14' X  11' high  with  a  door   8'x2.V'and2windows2J'x5'. 

b.  18' X 15' X 11' high  with  a  door  10' x3''^  and  4  windows  2i'x  5'. 

c.  20' X 18' X 12' high  with  a  door  11' x  3'   and  4  windows  2|'x  4'. 

d.  28' x  32' X 16' high  with  a  door  10' X  3'   and  4  windows  3'   x5'. 

e.  28' X  30' X  15' high  with  a  door  10' X  3'   and  3  windows  3'   x6'. 


CHAPTER   VI 
BUILDING  MATERIALS 

Besides  wood  many  materials  enter  into  the  construction  of 
buildings.  Among  these  materials  are  mortar,  cement,  stone, 
bricks,  marble,  slate,  etc. 

Mortar  is  a  paste  formed  by  mixing  lime  with  water  and 
sand  in  the  correct  proportions.  (Common  mortar  is  generally 
made  of  1  part  of  lime  to  5  parts  of  sand.)  It  is  used  to  hold 
bricks,  etc.,  together  and  when  stones  or  bricks  are  covered 
with  this  paste  and  placed  together,  the  moisture  in  the  mortar 
evaporates  and  the  mixture  "sets"  by  the  absorption  of  the 
carbon  dioxide  from  the  air.  Mortar  is  strengthened  by  add- 
ing cow's  hair  when  it  is  used  to  plaster  a  house  ;  in  such  mortar 
there  is  sometimes  half  as  much  lime  as  sand. 

Plaster  is  a  mixture  of  a  cheap  grade  of  gypsum  (calcium 
sulphate),  sand,  and  hair.  When  the  plaster  is  mixed  with 
water,  the  water  combines  with  the  gypsum  and  the  minute 
crystals  in  forming  interlace  and  cause  the  plaster  to  "  set." 

When  masons  plaster  a  house  they  estimate  the  amount  of 
work  to  be  done  by  the  square  yard.  Nearly  all  masons  use 
the  following  rule:  Calculate  the  total  area  of  walls  and  ceilings 
and  deduct  from  this  total  area  one  half  the  area  of  open- 
ings such  as  doors  and  windows.  A  bushel  of  mortar  will 
cover  about  3  sq.  yd.  with  two  coats. 

Example.  —  How  many  square  yards  of  plastering  are  neces- 
sary to  plaster  walls  and  ceiling  of  a  room  28'  by  32'  and  12'  high  ? 

Areas  of  the  front  and  back  walls  are  28  x  12  x    2  =    672  sq.  ft. 
Areas  of  the  side  walls  are  32  x  12  x    2  =    768  sq.  ft. 

Area  of  the  ceiling  is  28  x  32  =    896  sq.  ft. 

2336  sq.  ft. 

260  sq.  yd.  Ans. 
94 


CARPENTERING   AND   BUILDING  95 

EXAMPLES 

1.  What  will  it  cost  to  plaster  a  wall  10  ft.  by  13  ft.  at  $0.30 
per  square  yard  ? 

2.  What  will  it  cost  to  plaster  a  room  28'  6"  by  32'  4"  and 
9'  0"  high,  at  29  cents  a  square  yard,  if  one  half  their  area  is 
allowed  for  openings  and  there  are  two  doors  8'  by  3.V  and 
three  windows  6'  by  3'  3"  ? 

3.  What  will  it  cost  to  plaster  an  attic  22'  4"  by  16'  8"  and 
9'  4"  high,  at  a  cost  of  32  cents  a  square  yard  ? 

'  Bricks  used  in  Building 

Brickwork  is  estimated  by  the  thousand,  and  for  various 
thicknesses  of  wall  the  number  required  is  as  follows: 

8J-inch  wall,  or  1  brick  in  thickness,  14  bricks  per  superficial  foot ; 
12J-inch  wall,  or  1^  bricks  in  thickness,  21  bricks  per  superficial  foot; 
17-inch  wall,  or  2  bricks  in  thickness,  28  bricks  per  superficial  foot ; 
21^-inch  wall,  or  2\  bricks  in  thickness,  35  bricks  per  superficial  foot. 

EXAMPLES 
From  the  above  table  solve  the  following  examples : 

1.  How  much  brickwork  is  in  a  17"  wall  (that  is,  2  bricks 
in  thickness)  180'  long  by  6'  high  ? 

2.  How  many  bricks  in  an  8^"  wall,  164'  6"  long  by  6'  4"  ? 

3.  How  many  bricks  in  a  17"  wall,  48'  3"  long  by  4'  8"  ? 

4.  How  many  bricks  in  a  21^"  wall,  36'  4"  long  by  3'  6"  ? 

5.  How  many  bricks  in  a  12J"  wall,  38'  3"  long  by  4'  2"  ? 

6.  At  $19  per  thousand  find  the  cost  of  bricks  for  a  build- 
ing 48'  long,  31'  wide,  23'  high,  with  walls  12|"  thick.  There 
are  5  windows  (7'  x  3')  and  4  doors  (4'  x  8y)- 

To  estimate  the  number  of  bricks  in  a  wall  it  is  customary 
to  find  the  number  of  cubic  feet  and  then  multiply  by  22, 
which  is  the  numl)er  of  bricks  in  a  cubic  foot  with  mortar. 


96  VOCATIONAL  MATHEMATICS 


EXAMPLES 

1.  How  many  bricks  are  necessary  to  build  a  partition  wall 
36'  long,  22'  wide,  and  18"  thick  ? 

2.  How  many  bricks  will  be  required  for  a  wall  28'  6"  long, 
16'  8"  wide,  and  6'  5"  high  ? 

3.  How  many  cubic  yards  of  masonry  will  be  necessary  to 
build  a  wall  18'  4"  long  and  12'  2"  wide  and  4"  thick  ? 

4.  At  $19  per  thousand,  how  much  will  the  bricks  cost  to 
build  an  8f' ,  or  one-brick  wall,  28'  4"  long  and  8'  3" 
high? 

5.  At  $20.50  per  thousand,  how  much  will  the  bricks  cost  to 
build  a  12f "  wall,  52'  6"  long  and  14'  8"  high  ? 

6.  A  house  is  45'  x  34'  x  18'  and  the  walls  1  foot  thick,  the 
windows  and  doors  occupy  368  cu.  ft. ;  how  many  bricks  will 
be  required  to  build  the  house  ? 

7.  What  will  it  cost  to  lay  250,000  bricks,  if  the  cost  per 
thousand  is  $  8.90  for  the  bricks ;  $  3  for  mortar ;  laying,  $  8  ; 
and  staging,  $  1.25  ? 

Stone  Work 

Stone  work,  like  brick  work,  is  measured  by  the  cubic  foot 
or  by  the  perch  (16^'  X  1^'  X  1')  or  cord.  Practical  men  usu- 
ally consider  24  cubic  feet  to  the  perch  and  120  cubic  feet  to 
the  cord.     The  cord  or  perch  is  not  much  used. 

The  usual  way  is  to  measure  the  distance  around  the  cellar 
on  the  outside  for  the  length.  This  includes  the  corners  twice, 
but  owing  to  the  extra  work  in  making  corners  this  is  con- 
sidered proper.  No  allowance  is  made  for  openings  unless 
they  are  very  large,  when  one  half  is  deducted. 

The  four  walls  may  be  considered  as  one  wall  with  the 
same  height. 


CARPENTERING  AND  BUILDING  97 

Example.  —  If  the  outside  dimensions  of  a  wall  are  44'  by 
31',  10'  6"  high  and  8"  thick,  find  the  number  of  cubic  feet. 


25 


^^  ;^^  X  ^  X  4-  =  1060  cu.  ft.  Ans. 

160  ft.  length.  ^ 

Cement 

Some  buildings  have  their  columns  and  beams  made  of 
concrete.  Wooden  forms  are  first  set  up  and  the  concrete  is 
poured  into  them.  The  concrete  consists  of  Portland  cement, 
sand,  and  broken  stone,  usually  in  the  proportion  of  1  part 
cement  to  2  parts  sand  and  4  parts  stone.  The  average  weight 
of  this  mixture  is  150  pounds  per  cubic  foot.  After  the  con- 
crete has  "  set,"  the  wooden  boxes  or  forms  are  removed. 

Within  a  few  years  twisted  steel  rods  have  been  placed  in  the  forms 
and  the  concrete  poured  around  them.  This  is  called  reenforced  con- 
crete and  makes  a  stronger  and  safer  combination  than  the  whole  concrete. 
It  is  used  in  walls,  sewers,  and  arches.  It  takes  a  long  time  for  the  con- 
crete to  reach  its  highest  compressive  and  tensile  strength. 

Cement  is  also  used  for  walls  and  floors  where  a  waterproof  surface 
is  desired.  When  the  cement  "seta "  it  forms  a  layer  like  stone,  through 
which  water  cannot  pass.  If  the  cement  is  inferior  or  not  properly  made, 
it  will  not  be  waterproof  and  water  will  press  through  it  and  in  time 
destroy  it. 

EXAMPLES 

1.  If  one  bag  (cubic  foot)  of  cement  and  one  bag  of  sand  will 
cover  2\  sq.  yd.  one  inch  thick,  how  many  bags  of  cement  and 
of  sand  will  be  required  to  cover  30  sq.  yd.  2^"  thick  ? 

2.  How  many  bags  of  cement  and  of  sand  will  be  required 
to  lay  a  foundation  1"  thick  on  a  sidewalk  20'  by  8'  ? 

3.  How  many  bags  of  cement  and  of  sand  will  it  take  to 
cover  a  walk,  34'  by  8'  6",  }"  thick  ? 


98 


VOCATIONAL  MATHEMATICS 


4.  If  one  bag  of  cement  and  two  of  sand  will  cover  5^  sq.  yd, 
I"  thick,  how  much  of  each  will  it  take  to  cover  128  sq.  ft.  ? 

5.  How  much  of  a  mixture  of  one  part  cement,  two  parts 
sand,  and  four  parts  cracked  stone  will  be  needed  to  cover  a 
floor  28'  by  32'  and  8"  deep  ?     How  much  of  each  will  be  used  ? 

Shingles 

Shingles  for  roofs  are  figured  as  being  16"  by  4"  and  are  sold 
by  the  thousand.  The  widths  vary  from  2"  upwards.  They  are 
put  in  bundles  of  250  each.  When  shingles  are  laid  on  the 
roof  of  a  building  they  overlap  so  that  only  part  of  them  is 
exposed  to  the  weather. 

One  thousand  shingles  laid  4"  to  the  weather  will  cover 
100  square  feet,  or  ten  shingles  to  every  square  foot.  If  6" 
are  exposed  to  the  weather  600  shingles  will  be  necessary. 

1  square  =  100  square  feet. 


Table  for  Number  and  Weight  of  Pine  Shingles,  4"  Wide, 
FOR  One  Square  of  Roof 


Inches  exposed  to 

The   number  of  shingles 

the  weather  .     . 

4 

H 

5 

H 

6 

per  square  is  for  common 

Number  of  shingles 

gable  roofs.     For  hip  roofs 

for  one  square  of 

add  five  per  cent  to  these 

roof      .... 

1000 

800 

750 

6.55 

600 

figures. 

Weight    in    lb.    of 

The   weights    per   square 

shingles   on   one 

are   based   on   the   number 

square  of  roof    . 

240 

192 

180 

157 

144 

per    square :     1000    4-hich 
shingles  weigh  240  lb. 

There  are  several  methods  for  finding  the  number  of 
shingles  required  to  cover  a  roof.  One  is,  first  to  find  the 
number  of  squares  in  the  roof;  divide  this  number  by  1^ 
and  multiply  the  result  by  1000.^ 

1  This  rule  refers  to  shingles  laid  4J  inches  to  tlie  weather.  For  other 
couditions  refer  to  the  above  table. 


CARPENTERING  AND  BUILDING  99 

KxAMPLK. —  Find  the  number  of  shingles  required  to  cover 
a  roof  40  ft.  long  and  25  ft.  wide,  laid  4J^  inches  to  the  weather. 


40  X  26  =  1000  or  10  squares 
-  X  1000=10  X - 


*^  X  1000  =10  X  -  X  1000  =  8000.  Ans. 


Another  method,  used  when  the  roof  is  straight,  is  to  find 
the  number  of  courses  and  multiply  this  by  the  number  of 
shingles  in  a  course. 

Example.  —  The  ridge  of  a  roof  is  25  ft.  long  and  the 
rafters  are  15  ft.  on  each  side.  How  many  shingles  will  be 
required  to  cover  this  roof  ?  The  shingles  are  4''  wide.  Each 
shingle  is  5"  to  the  weather. 

26  -  J  =  75  30  -H  j%  =  72  76  x  72  =  6,400  Ans. 

This  gives  the  exact  nuinl>er  but  a  few  more  should  be  added  for 
waste,  etc.     Verify  answer  by  use  of  table  on  page  98. 

Another  method  is :  Since  it  takes  approximately  3  bunches 
(250  shingles  each)  laid  5"  to  the  weather  to  cover  a  square, 
multiplying  the  number  of  squares  in  the  roof  by  3  will  give 
the  number  of  bunches  required. 

Example.  —  If  a  roof  contains  50  squares,  how  many 
shingles  will  the  roof  need  to  cover  it  ? 

50  X  3  =  150  bunches  =  37,500  shingles,  Ans. 


EXAMPLES 

1.  How  much  will  it  cost  for  shingles  to  shingle  a  roof 
50  ft.  by  40  ft.,  if  1000  shingles  are  allowed  for  125  square 
feet  and  the  shingles  cost  $  1. 18  per  bundle  ? 

2.  Find  the  cost  of  shingling  a  roof  38  ft.  by  74  ft,  4"  to  the 
weather,  if  the  shingles  cost  $  1.47  a  bundle,  and  a  pound  and  a 
half  of  cuti  nails  at  $.0G  a  i)ound  are  used  with  each  bundle. 


100  VOCATIONAL  MATHEMATICS 

3.  How  many  shingles  would  be  needed  for  a  roof  having 
four  sides,  two  in  the  shape  of  a  trapezoid  with  bases  30  ft.  by 
10  ft.,  and  altitude  15  ft.,  and  two  (front  and  back)  in  the 
shape  of  a  triangle  with  base  20  ft.  and  altitude  15  ft.? 
(1000  shingles  will  cover  120  sq.  ft.) 

Slate  Roofing 

Slates  make  a  good-looking  and  durable  roof.  They  are  put 
on  with  nails  similarly  to  shingles.  Estimates  for  slate  roof- 
ing are  made  on  100  sq.  ft.  of  the  roof. 

The  following  are  typical  data  for  building  a  slate  roof  : 

A  square  of  No.  10  x  20  Monson  slate  costs  about  38.35, 
Two  pounds  of  galvanized  nails  cost  $.16  per  pound. 
Labor,  $S  per  square. 
Tar  paper,  at  2|  cents  per  pound,  1^  lb.  per  square  yard. 

EXAMPLES 

Using  the  above  data,  give  the  cost  of  making  slate  roofs 
for  the  following : 

1.  What  is  the  cost  of  laying  a  square  of  slate  ? 

2.  What  is  the  cost  of  laying  slate  on  a  roof  112'  by  44'  ? 

3.  What  is  the  cost  of  laying  slate  on  a  roof  156'  by  64'  ? 

4.  What  is  the  cost  of  laying  slate  on  a  roof  118'  by  52'  ? 

5.  What  is  the  cost  of  laying  slate  on  a  roof  284'  by  78'  ? 

Weight  of  Roof  Coverings 

Every  builder  should  know  the  weight  of  different  roof 
coverings  in  order  to  make  a  roof  strong  enough  to  support  its 
covering.  Besides,  allowance  must  be  made  for  ice  and  snow. 
The  weight  in  pounds  of  roof  covering  is  usually  expressed  in 
pounds  per  100  sq.  ft.,  or  square  of  roof. 


CARPENTERING   AND   BUILDING  101 


Approximate  WEUiiiT  ov  Roof  Covkrinos 

Na.mk                                                Wkioiit  pkr  1()0  bq.  pt. 

Sheathing,  Pine  1  inch  thick,  yellow  northern  .     .     .  300 

Sheathing,  Pine  1  inch  thick,  yellow  southern  .  400 

Spruce,  1  inch  thick 200 

Sheathing,  Chestnut  or  Maple,  I  inch  thick  ....  400 

Sheathing,  Ash,  Hickory,  or  Oak,  1  inch  thick  .     .     .  500 

Sheet  iron,  ^\  inch  thick 800 

Sheet  iron,  ^5  inch  thick,  and  laths 500 

Shingles,  Pine 200 

Slates,  \  inch  thick 1)00 

Skylights  (Glas-s,  f\  to  I  inch  thick) 250-700 

Sheet  Lead 500-800 

Cast  Iron  Plates,  |  inch  thick 1500 

Copper    80-126 

Felt  and  Asphalt 100 

Felt  and  Gravel 800-1000 

Iron,  Corrugated 100-2375 

Iron,  Galvanized  Flat    .     .     • 100-350 

Lath  and  Plaster 900-1000 

Thatch 650 

Tin 70-125 

Tiles,  Flat 1500-2000 

Tiles  (Grooves  and  Fillets) 700-1000 

Tiles,  Pan 1000 

Tiles,  with  Mortar 2000-3000 

Zinc 100-200 


EXAMPLES 
What  is  the  approximate  weight  of: 

1.  800  sq.  ft.  of  \  inch  slate  ? 

2.  1645  sq.  ft.  of  flat  tiles  ? 

3.  23.32  sq.  ft.  of  tiles  with  mortar? 

4.  3184  sq.  ft.  of  lath  and  plaster  ? 

5.  2789  sq.  ft.  of  sheathing  pine  1"  thick  ? 

6.  1841  sq.  ft.  of  pine  shingles  ? 


102  VOCATIONAL  MATHEMATICS 

7.  1794  sq.  ft.  of  thatch? 

8.  3279  sq.  ft.  of  felt  and  gravel  ? 

9.  1973  sq.  ft.  of  asphalt  ? 

10.    1589  sq.  "ft.  of  skylight  glass  i  in.  thick  ? 

Clapboards 

Clapboards  are  used  to  cover  the  outside  walls  of  frame 
buildings.  Most  clapboards  are  4  ft.  long  and  6  in.  wide.  They 
are  sold  in  bundles  of  twenty-five.  Three  bundles  will  cover 
100  square  feet  if  they  are  laid  4"  to  the  weather. 

To  find  the  number  of  clapboards  required  to  cover  a  given 
area,  find  the  area  in  square  feet  and  divide  by  IJ.  One 
quarter  the  area  should  be  deducted  to  allow  for  openings. 

EXAMPLES 

1.  How  many  clapboards  will  be  required  to  cover  an  area 
of  40  ft.  by  30  ft.  ? 

2.  How  many  clapboards  will  be  necessary  to  cover  an  area 
of  38'  by  42'  if  56  sq.  ft.  are  allowed  for  doors  and  windows  ? 

3.  How  many  clapboards  will  a  barn  60  ft.  by  50  ft.  require 
if  10%  is  allowed  for  openings  and  the  distance  from  founda- 
tion to  the  plate  is  17  ft.  and  the  gable  10  ft.  high  ? 

•;  *  ;        \ ;  :  y  r  \   '    Flooring 

Mcrst  floors  in  hou'seft,  are  made  of  oak,  maple,  birch,  or  pine, 
i^his  flooring  is  grooved  so  that  the  boards  fit  closely  together 
without  cracks  between  them. 

The   accompanying  figure  shows  the  ends  of     ^ ^ j— 1 

pieces  of  matched  flooring.     Matched  boards  are     ^ ^— '— ' 

also  used  for  ceilings  and  walls.  In  estimating  for  matched  flooring 
enough  stock  must  be  added  to  make  up  for  what  is  cut  away  from  the 
width  in  matching.  This  amount  varies  from  |"  to  |"  on  each  board  ac- 
cording to  its  size.     Some  is  also  wasted  in  squaring  ends,  cutting  up,  and 


CARPENTERING  AND  BUILDING  103 

fittinjj  to  exact  lengths.  A  common  floor  is  made  of  unmatched  boards 
and  is  usually  used  ;us  an  under  floor.  Not  more  than  \  is  allowed  for 
wastr. 

ExAMi'LK.  —  A  room  is  12  ft.  square  and  is  to  have  a  floor 
laid  of  matched  boards  IV'  wide;  one  third  is  to  be  added  for 
waste.  What  is  the  number  of  scjuare  feet  in  the  floor  ?  What 
is  the  number  of  board  feet  required  for  laying  the  floor? 

12  X  12  =  144  sq.  ft.  =  area.  144  x  J  =   48 

144  Ans.  144 

192  board  measure  for 

matched  floor. 
192  Ans. 
EXAMPLES 

1.  How  much  J  in.  matched  flooring  3"  wide  will  be  re- 
quired to  lay  a  floor  16  ft.  by  18  ft.  ?  One  fourth  more  is  al- 
lowed for  matching  and  3  %  for  squaring  ends. 

2.  How  much  hard  pine  matched  flooring  J"  thick  and  1^" 
wide  will  be  required  for  a  floor  13'  0"  X  14'  10"  ?  Allow  J  for 
matching  and  add  4  %  for  waste. 

3.  An  office  floor  is  10'  6"  wide  at  one  end  and  9'  6"  wide  at 
the  other  (trapezoid)  and  11'  7"  long.  What  will  the  material 
cost  for  a  maple  floor  |"  thick  and  1^"  wide  at  $60  per  M, 
if  4  sq.  ft.  are  allowed  for  waste  ? 

4.  How  many  square  feet  of  sheathing  are  required  for  the 
outside,  including  the  top,  of  a  freight  car  34'  long,  8'  wide, 
and  7^'  high,  if  12^%  is  allowed  for  waste  and  overhang? 

5.  In  a  room  50'  long  and  20'  wide  flooring  is  to  be  laid; 
how  many  feet  (board  measure)  will  be  required  if  the  stock 
is  J"  X  3"  and  \  allowance  for  waste  is  made? 

Stairs 

The  perpendicular  distance  between  two  floors  of  a  building 
is  called  the  rise  of  a  flight  of  stairs.  The  width  of  all  the 
steps  is  called  the  run.  The  perpendicular  distance  between 
steps  is  called  the  width  of  riser.     Nosing  is  the  slight  projec- 


104 


VOCATIONAL  MATHEMATICS 


tion  on  the  front  of  each 
step.  The  board  on 
each  step  is  the  tread. 

To  find  the  number  of 
stairs  necessary  to  reach 
from  one  floor  to  an- 
other: Measure  the  rise 
first.  Divide  this  by  8 
inches/  which  is  the  most 
comfortable  riser  for 
stairs.  The  run  should 
be  8i  inches  or  more 
to  allow  for  a  tread  of 
9|  inches  with  a  nosing  of  li  inches. 

Example.  —  How  many  steps  will  be  required,  and  what 
will  be  the  riser,  if  the  distance  between  floors  is  118  inches  ? 
118  --    8  =  14f  or  15  steps. 
118  -=-  15  =  7^1  inches  each  riser.    Ans. 


Stairs 


EXAMPLES 

1.  How  many  steps  will  be  required,  and  what  will  be  the 
riser,  (a)  if  the  distance  between  floors  is  8'  ?  (b)  If  the  dis- 
tance is  9  feet  ?  104 

2.  How  many  steps  will  be  required,  and  what  will  be  the 
riser,  (a)  if  the  distance  between  floors  is  12'  ?  (b)  If  the  dis- 
tance is  8'  8"  ? 

Carpenters'  Table  of  Wages 

To  find  the  amount  due  at  any  rate  from  30  cents  to  55  cents 
per  hour,  look  at  the  column  containing  the  rate  per  hour  and 
the  column  opposite  that  containing  the  number  of  hours,  and 
the  amount  will  be  shown.  Time  and  a  half  is  counted  for 
overtime  on  regular  working  days,  and  double  time  for  Sundays 
and  holidays. 

1  Other  distances  may  also  be  used  if  required. 


CARPENTERING   AND   BUILDING 


105 


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EXAMPLEvS 

1.  Find  the  amount  due  a  carpenter  who  has  worked  8  hours 
regular  time  and  2  hours  overtime  at  55  cents  per  hour. 

2.  A  carpenter  worked  on  Sunday  from  8  to  11  o'clock.  If  his 
regular  wages  are  45  cents  per  hour,  how  much  will  he  receive  ? 

3.  A  carpenter  received  55  cents  an  hour.  How  much  money 
is  due  him  for  working  July  4th  from  8-12  a.m.  and  1-4.30  p.m.  ? 

4.  A  carpenter  works  six  days  in  the  week ;  every  morning 
from  7.30  to  12  m.  ;  three  afternoons  from  1  to  4.30  p.m.  ;  two 
afternoons  from  1  to  5.30 ;  and  one  from  1  until  6  p.m.  What 
will  he  receive  for  his  week's  wage  at  50  cents  per  hour  ? 


106  VOCATIONAL    MATHEMATICS 

Painting 

Paint,  which  is  composed  of  dry  coloring  matter  or  pigment  mixed 
with  oil,  drier,  etc.,  is  applied  to  the  surface  of  wood  by  means  of  a 
brush  to  preserve  the  wood.  The  paint  must  be  composed  of  materials 
which  will  render  it  impervious  to  water,  or  rain  would  wash  it  from  the 
exterior  of  houses.  It  should  thoroughly  conceal  the  surface  of  whatever 
it  is  applied  to.  The  unit  of  painting  is  one  square  yard.  In  painting 
wooden  houses  two  coats  are  usually  applied. 

It  is  often  estimated  that  one  pound  of  paint  will  cover  4  sq.  yd.  for 
the  first  coat  and  6  sq.  yd.  for  the  second  coat.  Some  allowance  is  made 
for  openings  ;  usually  about  one  half  of  the  area  of  openings  is  deducted, 
for  considerable  paint  is  used  in  painting  around  them. 

Table 

1  gallon  of  paint  will  cover  on  concrete      .     .    300  to  375  superficial  feet 

1  gallon  of  paint  will  cover  on  stone  or  brick 

work 190  to  225  superficial  feet 

1  gallon  of  paint  will  cover  on  wood  .     .     .     .     375  to  525  superficial  feet 

1  gallon  of  paint  will  cover  on  well-painted  sur- 
face or  iron 600  superficial  feet 

1  gallon  of  tar  will  cover  on  first  coat     ...  90  superficial  feet 

1  gallon  of  tar  will  cover  on  second  coat    .     .  160  superficial  feet 

EXAMPLES 

1.  How  many  gallons  of  paint  will  it  take  to  paint  a  fence 
6'  high  and  50'  long,  if  one  gallon  of  paint  is  required  for 
every  350  sq.  ft.  ? 

2.  What  will  the  cost  be  of  varnishing  a  floor  22'  long  and 
16'  wide,  if  it  takes  a  pint  of  varnish  for  every  four  square 
yards  of  flooring  and  the  varnish  costs  $  2.65  per  gal.  ? 

3.  What  will  it  cost  to  paint  a  ceiling  36'  by  29'  at  21  cents 
per  square  yard  ? 

4.  What  will  be  the  cost  of  painting  a  house  which  is  52' 
long,  31'  wide,  21'  high,  if  it  takes  one  gallon  of  paint  to  cover 
300  sq.  ft.  and  the  paint  costs  $1.65  per  gallon  ? 


PART   III  — SHEET   AND   ROD   METAL   WORK 

CHAPTER   VII 
Blanking  or  Cutting  Dies 

Many  kinds  of  receptacles  are  pressed  or  cut  from  difTerent 
kinds  of  sheet  metal,  such  as  copper,  tin,  and  aluminum.  Cans, 
pots,  parts  of  metal  boxes,  and  all  sorts  of  metal  novelties  are 
punched  out  of  sheet  metal  most  economically  by  the  punch 
and  die  operated  by  the  hydraulic  press.  So  skillfully  can  die 
makers  produce  dies  and  punches  to  cut  out  articles  that  thou- 
sands of  everyday  necessities  in  the  household  are  made  by 
this  method.  Parts  of  watches,  parts  of  automobiles,  and  parts 
of  machinery  are  punched  out.     Some  presses  that  operate  the 


Blanking"  or  "Cutting"  Diks 


Cutting  dies  consist  of  an  upper  "  male"  die  or  "punch,"  and  the  lower 
or  "female"  die.  Circumstances  determine  whether  any  or  how  much 
"shear"  shall  be  given  to  the  cutting e<Ige.  For  ordinary  work  in  tin,  brass, 
etc.,  a  moderate  amount  of  shear  is  desirable.  Ordinarily,  the  steel  rutting 
rings  are  welded  to  wrought-iron  plates,  after  which  they  are  hardened,  care- 
ftilly  tempered,  and  ground  on  s|>ecial  nnu-hinery.  In  some  rases  it  is  prefer- 
able to  fasten  the  steel  dies  in  cast-iron  chucks  or  die-beds  by  means  of  keys 

107 


108  VOCATIONAL   MATHEMATICS 


or  screws.  This  applies  more  particularly 
to  small  dies.  Cutting  dies  may  be  made 
to  tit  any  size  and  style  of  press.  For 
cutting  thick  iron,  steel,  brass,  and  other 
heavy  metals,  both  the  die  and  punch 
should  be  hard  and  provided  with  strip- 
pers. 


Punch  and  Die  without  Stripper  Punch  and  Die  with  Strippkr 

punch  and  die  are  run  by  foot  power,  but  those  most  generally 
used  are  run  by  electricity. 

A  blanking  or  cutting  die  is  a  metal  plate  or  disk  having 
an  opening  in  the  center  used  in  a  punching  machine  or  press 
which  is  supplied  with  ample  power  and  which  also  supports 
the  metal  from  which  pieces  are  punched.  Dies  are  made  in 
almost  any  size  and  shape  for  cutting  flat  blanks  in  tin,  iron, 
steel,  aluminum,  brass,  copper,  zinc,  silver,  paper,  leather,  etc. 

Holes  are  punched  in  thick  sheets  of  metal  or  in  heavy 
plates  by  means  of  great  pressure  exerted  by  a  hydraulic  press. 
This  pressure,  in  pounds,  is  usually  about  60,000  times  the 
area  (expressed  in  square  inches)  of  the  surface  cut  out,  or  in 
other  words  about  60,000  lb.  per  square  inch. 

EXAMPLES 

1.  How  much  pressure  will  be  necessary  for  a  hydraulic 
press  to  exert  on  a  sheet  of  boiler  plate  ^y  thick,  if  it  is  de- 
sired to  punch  holes  ^"  in  diameter  ? 

2.  How  much  pressure  will  it  be  necessary  to  use  in  order 
to  punch  holes  f"  in  diameter  out  of  \"  boiler  plate  ? 

3.  How  much  pressure  will  it  be  necessary  to  use  in  order  to 
punch  jV  holes  out  of  j^"  boiler  plate? 


SHEET    AND    HOD    METAL   WORK 


109 


Combination  Dies 

Double  dies  for  blanking;  and  perforating  are  extensively 
used  in  the  manufacture  of  washers,  key  blanks,  electrical 
instruments,  hardware,  etc.  The  blanking  and  perforating 
punches  act  simultaneously.     At  a  single  stroke  of  the  ])ress 


Double"  Dies  for  Blankino  and  Perforating 


Tlie  perforating  and  blanking  punches  act  simultaneously  in  such  a  manner 
that  the  lioles  are  punched  first,  whereupon  the  strip  or  slieet  being  fed  for- 
ward, the  blank  is  cut  out  around  them  and  the 
holes  for  the  next  blank  perforate*!  at  the  same   /" 
stroke.     In  this  manner  a  blank  is  <         '  "  ' 
forated  at  every  stroke  of  tlie  press. 

The  same  principle  may  be  extended  so  as  to 
punch  a  number  of  perforated  blanks  at  a  time. 


ate<l  at  tne  same   /    >— ^  \        ^^^ 

is  completely  per-      f      }  O  )  O  (  O  ) 


Perfobatino  Dibs  with  Stripper  Plates 


110  VOCATIONAL   MATHEMATICS 

one  punch  perforates  the  holes  and  the  other  punch  cuts  out 
the  metal  between  the  holes  punched  at  the  previous  stroke. 
These  dies  are  usually  so  arranged  that  the  finished  article  is 
automatically  pushed  out  from  the  dies  by  the  action  of  the 
springs.  An  expert  operator  can  punch  many  thousands  of 
pieces  in  a  day. 

Example.  —  How  much  metal  will  be  required  for  2000 
blacking  box  covers,  6"  in  diameter,  and  f  deep  ? 

6"  +  f "  +  f"  =  7i",  diameter  of  one  cover. 
.7854  X  7.5  X  7.5  =  44.1787  sq.  in.,  area. 

44.1787  X  2000  =  88,357.4  sq.  in.  =  613.6  sq.  ft.     Ans. 

EXAMPLES 

1.  How  large  must  the  blank  be  cut  for  a  pail  cover  that  is 
to  be  (a)  5"  in  diameter  and  |"  deep  ?  (6)  7"  in  diameter  and 
ly  deep  ?  (c)  6"  in  diameter  and  V'  deep  ?  (d)  8"  in  diameter 
and  li"  deep  ? 

2.  How  large  must  the  blank  be  cut  for  the  pail  bottom  in 
each  of  the  above  examples  (a,  b,  c,  and  d)  ? 

The  blank  must  be  i"  larger  in  diameter  than  the  diameter  of  the  part 
in  order  to  allow  |"  all  around  for  forming  into  the  sides. 

3.  How  large  must  the  piece  be  to  form  the  sides  of  the 
pail  in  each  of  the  above  examples  ?  Pail  (a)  to  be  6"  high, 
(6)  9"  high,  (c)  8"  high,  and  (d)  10''  high. 

Allow  ^"  in  height,  and  for  the  lock  seam  allow  f"  in  circumference. 

4.  How  much  metal  will  be  necessary  for  850  complete  pails 
in  example  (a)  above  ? 

5.  How  much  metal  will  be  necessary  for  1500  complete 
pails  in  example  (6)  above  ? 

6.  How  much  metal  will  be  necessary  for  840  complete  pails 
in  example  (c)  above  ? 

7.  How  much  metal  will  be  necessary  for  1000  complete 
pails  in  example  (d)  above  ? 


SHEET   AND    ROD    METAL   WORK 


111 


Ma<him:  i<>h   I  >i.\«;  Dies 


8.  How  many  square  feet  of  sheet  copper  will  be  required 
to  make  a  rectangular  tank  7'  long,  3'  wide,  IV  deep,  allowing 
10  <Jo  extra  for  waste  ?    The  tank  is  to  be  open  on  to}). 


112  VOCATIONAL   MATHEMATICS 

9.  If  the  sheet  copper  weighs  12  lb.  per  sq.  ft.  and  costs 
25  cents  per  lb.,  what  will  be  the  cost  of  the  material  used? 

10.  Find  the  amount  of  material  in  a  straight  piece  of  copper 
pipe  24"  diameter  and  8'  long.  The  pipe  is  formed  from  sheet 
copper  weighing  15.4  lb.  per  sq.  ft.     What  does  it  weigh  ?  ■ 

Allowance  for  seam,  1\". 

11.  If  the  cost  of  copper  in  Example  10  is  27  cents  per  lb. 
and  the  amount  of  copper  wasted  iu  making  the  pipe  is  equal 
to  ^  the  finished  weight  of  the  pipe,  what  is  the  total  cost  of 
material  used  ? 

12.  What  is  the  cost  per  pound  of  finished  pipe  for  the  ma- 
terial used  in  Examples  10  and  11  ? 

13.  If  the  labor  to  make  the  pipe  in  Example  10  costs  $  36, 
what  is  the  cost  per  lb.  for  labor  ? 

14.  What  will  be  the  total  cost  in  Example  10  of  all  mate- 
rial and  labor  per  pound  of  finished  pipe  ? 

Circles 

Review  the  paragraphs  on  the  Circle  on  pages  62  and  63,  and 
on  the  Cylinder,  page  72. 

EXAMPLES 

To  find  circumference  of  a  cylindrical  tank,  when  the  diameter 
is  given. 

1.  What  is  the  circumference  of  a  tin  plate  (a)  8"  in  diame- 
ter ?  (b)  14"  in  diameter  ?  (c)  V  6"  in  diameter  ?  (d)  2'  3" 
in  diameter  ?     (e)  1'  9"  in  diameter  ? 

2.  What  is  the  diameter  of  a  copper  plate  containing,  (a) 
25  sq.  in.  ?  (b)  43  sq.  ft.  ?  (c)  349  sq.  in.  ?  (d)  8840  sq.  in.  ? 
(e)  616  sq.  in.  ?     (/)  340  sq.  in.? 

3.  What  is  the  diameter  of  an  iron  tank  (a)  6'  high,  contain- 
ing 40  gallons?  (b)  5^'  high,  containing  360  gallons?  (c)  6' 
high,  containing  120  gallons  ?     (il)  12'  high,  containing  141  gal- 


SHEET    AND    ROD    METAL   WORK  113 

Ions  ?     (e)  8'  high,  containing  181  gallons  ?     (/)  4'  high,  con- 
taining 241  gallons  ?     (231  cu.  in.  =  1  gal.) 

Two  circles  are  formed  by  a  plane  passing  through  a  hollow  cylinder  of 
metal,  as  it  has  an  outside  and  an  inside  circumference,  due  to  the  thick- 
ness of  the  metal.  A  circle  the  circumference  of  which  is  midway 
between  these  two  circumferences  is  said  to  be  the  neutral  circle  or  ring, 
and  its  diameter  is  the  neutral  diameter. 

To  illustrate  :  the  neutral  diameter  of  a  ring  8"  outside  diameter  and 
6"  inside  diameter  is  7". 

4.  What  is  the  length  of  a  smoke  arch  ring  whose  outside 
diameter  is  60"  and  a  section  of  which  measures  2^"  x  2.J^"  ? 

The  length  of  the  ring  is  the  length  or  circumference  of  a  circle  whose 
dianieter  is  the  neuti-al  diameter  of  the  ring. 

5.  A  smoke  stack  32"  inside  diameter  is  made  from  ^^^  iron 
sheet  stock.  What  would  be  the  length  of  the  sheet,  allowing 
2"  for  lapping? 

Use  the  neutral  diameter. 

6.  If  we  desire  to  make  a  close  fit  over  the  above  sheet  and 
allow  2"  for  overlapping,  what  size  sheet  will  we  use  ? 

7.  A  blacksmith  desires  to  place  a  band  around  a  hub  that 
is  17"  in  diameter.  If  the  band  is  2"  square  and  if  no  allow- 
ance is  made  for  shrinking,  how  long  a  piece  of  iron  is  necessary 
to  do  the  job? 

8.  If  the  above  band  averages  ^  lb.  per  cu.  in.,  what  is  the 
weight  of  the  band  ? 

9.  A  flat  iron  ring  casting  has  the  following  dimensions : 
18"  outside  diameter,  and  9"  inside  diameter,  and  3"  thictk. 
What  will  it  cost  at  3  cents  a  pound?  What  is  the  cost  of 
a  cubic  foot  of  the  iron  ?     (1  cu.  in.  iron  =  .26  lb.) 

10.  The  neutral  diameter  of  an  engine  wheel  tire  is  76". 
If  the  steel  weighs  .28  lb.  per  cu.  in.,  what  would  be  the 
weight  of  the  foregoing,  if  its  section  was  a  trapezoid  whose 
dimensions  were  6"  wide  on  the  face,  3J"  thick  on  one  side, 
and4J"  on  the  other? 


114  YOGATIONAL   MATHEMATICS 

11.  What  will  be  the  weight  per  sq.  ft.  of  a  steel  plate  |" 
thick  if  a  steel  plate  \"  thick  weighs  20  lb.  per  sq.  ft.  ? 

12.  What  is  the  weight  of  a  steel  bar  11'  long  having  a  cross 
section  area  of  6|  sq.  in.  ?     (1  cu.  ft.  steel  =  490  lb.) 

13.  A  "  1226  class  "  draft  pan  sheet  was  laid  out  in  the  form 
of  a  rectangle,  the  length  of  which  measured  Q2^"  and  the 
width  22".     Find  its  area  in  square  feet. 

14.  If  the  above  sheet  were  \"  thick  and  weighed  9  lb.  per 
sq.  ft.,  what  would  be  its  weight  after  deducting  7  lb.  for 
rivet  holes  ? 

15.  An  iron  plate  is  divided  into  four  sections;  the  first 
contains  29J  sq.  in.,  the  second  contains  50|  sq.  in.,  the  third 
contains  41  sq.  in.,  and  the  fourth  69y^g-  sq.  in.  How  many- 
square  inches  iu  the  plate  ?     If  2"  thick,  what  does  it  weigh  ? 

Calculate  480  lb.  to  a  cubic  foot. 

16.  What  is  the  area  of  the  surface  of  a  boiler  plate  3'  8" 
by  1'  6"  ? 

17.  How  many  square  pieces  of  zinc  6"  by  &'  can  be  cut 
from  a  zinc  plate  3'  by  6'  ? 

18.  How  many  pieces  of  zinc  4"  by  8"  can  be  cut  from  a 
zinc  plate  3'  by  6'  ? 

19.  What  is  the  value  of  the  copper  in  a  copper  tank  measur- 
ing 4|'  X  3'  6"  X  2'  3"  and  made  of  copper  weighing  12  lb.  per. 
sq.  ft.,  if  the  copper  costs  25  cents  per  lb.  and  no  account  is 
made  of  laps  and  seams  and  waste?     Tank  open  at  top. 

20.  If  a  sheet  of  copper  30"  by  60"  weighs  25  lb.,  what  is 
its  weight  per  sq.  ft.  ? 

21.  What  would  be  the  length  of  a  40  lb.  sheet  of  the  same 
thickness  of  copper,  16"  wide  ? 

Use  the  result  from  Example  20. 

22.  What  is  the  capacity  of  a  cylindrical  tank  9'  long  and  5' 
diameter,  inside  measurements? 


SHEET   AND    ROD   METAL   WORK  115 

23.  What  would  be  the  dimensions  of  a  cubical  tank  contain- 
ing the  same  amount? 

24.  On  a  base  4'  x  5',  how  high  shall  a  tank  be  made  to  con- 
tain 14  tons  of  water  which  weighs  64  lb.  per  cu.  ft.  ? 

25.  How  many  sq.  ft.  of  sheet  metal  will  be  required  to  form 
the  bottom  and  sloping  sides  of  a  pan,  the  bottom  2}/'  square, 
the  slant  height  of  the  sides  8",  and  the  opening  across  the 
open  top  3\"  square  ?  Do  not  consider  any  allowances  for 
waste,  laps,  etc. 

26.  What  is  the  area  of  one  side  of  a  sheet  ring  that  is  3 
inches  inside  diameter  and  4  inches  outside  diameter  ? 

27.  The  circular  basin  of  a  washing  machine  is  20  feet  in 
diameter.  What  will  it  cost  to  copper  line  the  bottom  at  $  1.20 
per  sq.  ft.  ? 

28.  A  piece  of  steel  shafting  10'  3"  long  is  rough  machined 
to  a  diameter  of  11  J".  When  finished  to  a  diameter  of  11^"  and 
with  a  1"  diameter  hole  running  through  its  entire  length,  how 
much  has  it  been  reduced  in  weight? 

29.  If  the  average  weight  of  wrought  iron  is  480  lb.  per 
cu.  ft.,  what  is  the  weight  of  a  piece  of  bar  iron  1"  square,  9' 
long?  What  would  be  the  area  of  the  cross  section  of  a  bar  6' 
long  weighing  72  lb.  ? 

30.  A  solid  cast  iron  cone  pulley  is  3'  2^"  long.  The  diame- 
ter of  one  end  of  the  pulley  is  4J"  and  the  other  end  of  the 
]mlley  is  lOV'-  A  hole  2"  in  diameter  runs  through  the  entire 
length  of  the  pulley.  What  is  its  weight  ?  (Cast  iron  weighs 
450  lb.  per  cu.  ft.) 

31.  How  many  square  feet  of  sheet  copper  will  be  required 
to  make  a  rectangular  tank  7'  long,  3'  wide,  IJ"  deep,  allowing 
10  %  extra  for  waste  ?     Tank  open  on  top. 

32.  If  the  sheet  copper  weighs  12  lb.  per  sq  ft.  and  costs  29 
cents  per  lb.,  what  will  be  the  cost  of  the  material  used  ? 


116 


VOCATIONAL   MATHEMATICS 


33.  The  average  weight  of  wrought  iron  is  480  lb.  per  cu.  ft. 
A  bar  4"  square  and  3'  long  weighs  how  many  pounds  ? 

34.  A  cast  iron  pulley  is  4'  3^"  long.  The  diameter  of  one 
end  of  the  pulley  is  5f "  and  of  the  other  end  of  the  pulley  is 
11  J".  A  hole  If"  in  diameter  runs  through  the  entire  length 
of  the  pulley.     What  is  its  weight  ? 

35.  How  many  bosom  pieces  15J"  long  can  be  made  from  a 
bar  of  angle  iron  25'  long,  and  how  much  waste  will  there  be? 

36.  How  many  clips  9y^^"  long  can  be  made  from  a  bar  25' 
long,  and  what  will  be  the  waste  ? 


Standard  Gauge  for  Sheet  Metal  and  Wire 

In  order  to  measure  the  thickness  of  a  piece  of  wire  or  sheet 
metal,  a  gauge  has  been  made.  The  U.  S.  standard  gauge  is  a 
circular  instrument  S\"  in  diameter  and  about  i"  thick.     The 

gauge  numbers,  which 
run  from  0  to  36,  are 
those  adopted  by  Con- 
gress, March  3,  1893. 
The  gauge  is  made  of 
hardened  and  tempered 
steel. 

To  measure  wire  or 
sheet  metal,  the  wire 
or  metal  is  placed  in  a 
perforation  that  it  fits, 
and  the  number  of  this 
perforation  is  called  the 
number  of  the  gauge 
of  the  wire  or  metal. 
To  find  the  size  of  a 
wire  or  piece  of  sheet  metal  in  decimal  parts  of  an  inch  when 
the  gauge  has  been  determined,  the  following  table  should  be 
used: 


U.  S.  Standard  Gauge 


SHEET   AND    HOD    METAL    WORK 


117 


StANI)AR1>S   for   WiKK  (iAHOK 
Dimensions  ofSiKos  in  Decinml  Parts  of  an  Inch 


BlRMINO- 

Nl-MBKR 

AmkRH  AN 

IIAM,  OR 

Waphhi'rn 

Impkriai. 

Stubs 

U.S. 

Nl'MBKR 

or  WiRK 

OR  Br«>wn 

Stibi* 

*  MOEN 

WlKK 

Stkbi. 

Stani>ari» 

«>K  VV'iKK 

(Jakje 

iS^  SlIARPK 

Iron 

WiBB 

Mr.j.  Co. 

Galgk 
.464 

WiRB 

KoR  Plate 

itAvr.y. 

000000 

.46875 

000000 

00000 

.... 

.... 

.... 

.432 

.... 

.4375 

00000 

0000 

.46 

.454 

.3938 

.400 

.40625 

0000 

000 

.40964 

.425 

.3625 

.372 

.375 

000 

00 

.3648 

.38 

.3310 

.348 

.34375 

00 

0 

.32486 

.34 

.3065 

.324 

.... 

.3125 

0 

1 

.2893 

.3 

.2830 

.300 

.227 

.28125 

1 

2 

.25763 

.284 

.2625 

.276 

.219 

.265625 

2 

3 

.22942 

.259 

.2437 

.252 

.212 

.25 

3 

4 

.20431 

.238 

.2253 

.232 

.207 

.234375 

4 

5 

.18194 

.22 

.2070 

.212 

.204 

.21875 

5 

6 

.16202 

.203 

.1920 

.192 

.201 

.203125 

6 

7 

.14428 

.18 

.1770 

.176 

.199 

.1875 

7 

8 

.12849 

.165 

.1620 

.160 

.197 

.171875 

8 

9 

.11443 

.148 

.1483 

.144 

.194 

.15625 

9 

10 

.10189 

.134 

.1350 

.128 

.191 

.140625 

10 

11 

.090742 

.12 

.1205 

.116  . 

.188 

.125  — 

11 

12 

.080808 

.109 

.1055 

.104 

.185 

.109375 

12 

13 

.071961 

.095 

.0915 

.092 

.182 

.09375 

13 

14 

.064084 

.083 

.0800 

.080 

.180 

.078125 

14 

15 

.057068 

.072 

.0720 

.072 

.178 

.0703125 

15 

16 

.05082 

.065 

.0625 

.064 

.175 

.0625 

16 

17 

.045257 

.058 

.0540 

.056 

.172 

.05625 

17 

18 

.040303 

.049 

.0475 

.048 

.168 

.05 

18 

19 

.03589 

.042 

.0410 

.040 

.164 

.04375 

19 

20 

.031961 

.035 

.0348 

.036 

.161 

.0375 

20 

21 

.028462 

.032 

.03175 

.032 

.157 

.034375 

21 

22 

.025347 

.028 

.0286 

.028 

.155 

.03125 

22 

23 

.022571 

.025 

.0258 

.024 

.153 

.028125 

23 

24 

.0201 

.022 

.0230 

.022 

.151 

.025 

24  > 

26 

.0179 

.02 

.0204 

.020 

.148 

.021875 

25 

26 

.01594 

.018 

.0181 

.018 

146 

.01875 

26 

27 

.014195 

.016 

.0173 

.0164 

.143 

.0171875 

27 

28 

.012641 

.014 

.0162 

.0149 

.139 

.015625 

28 

29 

.011257 

.013 

.0150 

.01.36 

.134 

.0140625 

29 

30 

.010025 

.012 

.0140 

.0124 

.127 

.0125 

30 

31 

.008928 

.01 

.0132 

.0116 

.120 

.0109375 

31 

32 

.00795 

.009 

.0128 

.0108 

.115 

.01015625 

32 

33 

.00708 

.008 

.0118 

.0100 

.112 

.009375 

33 

34 

.006304 

.007 

.0104 

.0092 

.110 

.00859375 

34 

35 

.005614 

.005 

.0095 

.0084 

.108 

.0078125 

34 

36 

.005 

.004 

.0090 

.0076 

.106 

.00703125 

36 

37 

.004453 

.... 

.... 

.0068 

.103 

.006640625 

37 

38 

.003965 

.0060 

.101 

.00625 

38 

39 

.003531 

.... 



.0052 

.099 

39 

40 

.003144 





.0048 

.097 



40 

The  American  or  B.  «k  8.  frauge  l«  the  standard  for  sheet  brass,  copper,  or  German  silver, 
and  for  wire  of  the  same  material. 

The  Binnin^'ham  (range  is  use<l  for  soft  iron  wire  or  rods. 

The  Washburn  A  Moen  gauge  is  used  for  iron  or  copper  telegraph  and  telephone  wire. 

Stubs  Steel  Wire  gauge  is  the  standard  for  Stubs  drill  rods.  It  is  not  the  same  as  Stubs 
Iron  Wire  gauge. 

The  U.  S.  Sundard  gauge  is  recognized  as  standard  for  sheet  Irou  and  steel. 


118 


VOCATIONAL   MATHEMATICS 


EXAMPLES 

1.  Find  the  thickness  of  wire  of  (a)  No.  10  gauge;  (6)  No.  28 
gauge ;  (c)  No.  15  gauge ;  (d)  No.  5  gauge ;  (e)  No.  3  gauge. 

2.  Find  the  thickness  of  sheet  metal  No.  7  gauge. 

3.  Find  the  thickness  of  iron  of  (a)  No.  24  gauge ;  (b)  No. 
18  gauge. 

4.  Find  the  thickness  of  steel  of  (a)  No.  29  gauge;  (b) 
No.  16  gauge. 

5.  Find  the  weight  of  120  sq.  ft.  of  (a)  No.  4  sheet  iron ; 
(6)  28  sq.  ft.  of  No.  19  sheet  steel ;  (c)  35  sq.  ft.  of  No.  16 
sheet  iron;  (d)  79  sq.  ft.  of  No  5  sheet  steel. 

See  page  120  and  the  tables  on  pages  121-123. 

6.  Find  the  weight  of  38J  sq.  ft.  of  sheet  steel  whose  thick- 
ness is  .125. 

7.  Find  the  weight  of  69^  sq.  ft.  of  sheet  iron  whose  thick- 
ness is  .04375. 

8.  Find  the  weight  of  1281  gq.  ft.  of  No.  8  sheet  iron. 

9.  Find  the  weight  of  250^  sq.  ft.  of  No.  6  sheet  steel. 

10.  What  number  gauge  wire  (B.  &  S.)  is  .090742  ? 

11.  What  number  gauge  wire  (Stubs  Steel  Wire)  is  .157  ? 

Additional  Tables  for  Sheet  Metal  Workers 

Tin  Plate 


Thickness  Stub's 
Gauge 

No.  Sheets  in  Box 

Net  Weight  ok 
Box  14  X  20  Sheets 

Taggers 

38  (34) 

225  (150) 

112  lb. 

IC 

30 

112 

107  lb. 

IX 

28 

112 

135  lb. 

IXX 

27 

112 

I3i)  lb. 

IXXX 

26 

112 

176  lb. 

IXXXX 

25 

112 

1-96  lb. 

SHEET   AKD    ROD    METAL   WORK 


110 


Sheet  Zinc  —  M.  &  H.  Gauge 


No. 

1  =  0.002  in. 

No.  11  =0.024  in. 

No.  21  =  0.080  in. 

No. 

2  =  0.004  in. 

No.  12  =  0.028  in. 

No.  22  =  0.090  in. 

No. 

S  =  0.m\  in. 

No.  13  =  0.0:^2  in. 

No.  23  =  0.100  in. 

No. 

4  =  0.008  in. 

No.  14  =  0.0:J6  in. 

No.  24  =0.125  in. 

No. 

5  =  0.010  in. 

No.  15  =  0.040  in. 

No.  25  =  0.250  in. 

■  No. 

6  =  0.012  in. 

No.  16  =  0.045  in. 

No.  26  =  0.375  in. 

No. 

7  =  0.014  in. 

No.  17  =  0.050  in. 

No.  27  =  0.500  in. 

No. 

8  =  0.010  in. 

No.  18  =  0.065  in. 

No.  28  =  1.000  in. 

No. 

9  =  0.018  in. 

No.  19  =  0.060  in. 

No. 

10  =  0.020  in. 

No.  20  =  0.070  in. 

American  Ucssia  Iron 


No.  7  =  .015  in. 

No.  12  =  .021  in. 

No.  8  =  .016  in. 

No.  13  =  .024  in. 

No.  9  =  .017  in. 

No.  14  =  .025  in. 

No.  10  =  .018  in. 

No.  15  =  .027  in. 

No.  11  =.020  in. 

No.  16  =  .030  in. 

1.  AVhat  is  the  net  weight  of  4  boxes  tin  plate  IC  ? 

2.  What  is  the  net  weight  of  8  boxes  tin  plate  IXX  ? 

3.  What  is  the  net  weight  of  5  boxes  tin  plate  IXXXX? 

4.  What  is  the  net  weight  of  7  boxes  tin  plate  IXXX  ? 

5.  What  is  the  net  weight  of  12  boxes  tin  plate  IXX  ? 

6.  How  many  sheets  in  6  boxes  tin  plate  IXX  ? 

7.  How  many  sheets  in  8  boxes  tin  plate  IXXXX  ? 

8.  How  many  sheets  in  7  boxes  tin  plate  IXXX  ? 

9.  How  many  sheets  in  10  boxes  tin  plate  IX  ? 
10.  How  many  sheets  in  14  boxes  tin  plate  IC  ? 


120  VOCATIONAL   MATHEMATICS 

Weights  and  Areas 

Iron  and  steel  bars  are  sold  in  round,  square,  or  hexagonal 
shape.  When  it  is  necessary  to  know  the  area  or  weight  of  a 
bar,  look  at  the  left  column  in  tables  similar  to  the  following 
for  the  number  corresponding  to  the  thickness  or  diameter  of 
the  bar,  then  in  the  same  line  in  the  columns  to  the  right  the 
weight  or  area  is  found. 

Example.  —  What  is  the  weight  of  a  round  bar  of  steel  12' 
long  and  j\"  in  diameter  ? 

According  to  the  table  on  page  122,  the 

Wt.  of  ^"  bar  per  in.  length  =  .0218  lb. 
Wt.  of  j%"  bar  per  ft.  length  =  .2616  lb. 
Wt.  of  t\"  bar  12  ft.  in  length  =  3.1392  lb.     Ans. 

To  obtain  the  weight  of  a  certain  size  sheet  steel  or  iron  one 
should  look  in  the  table  for  the  weight  per  square  foot  corre- 
sponding to  the  size  that  it  is  desired  to  know  and  when  the 
weight  per  square  foot  is  known,  any  number  of  feet  can  be 
found  by  multiplying  the  weight  per  foot  by  the  number  of 
feet. 


SHEET  AND  ROD  METAL  WORK 


121 


Thickness  and  Weight  of  Sheet  Steel  and  Ieon 

Adopted  by  U.  S.  Oovemment  July  1,  1893 

Weight  of  1  cu.  ft.  is  assamed  to  bo  487.7  lb.  for  steel  plates  and  480  lb.  for  iron  plates. 


NrMBRS  OF 

AppSOXIMATK  TniCKNBM 

Wkioiit  pkr  Sy.  Ft. 

OVBBWEtOIlT 

Gatuk 

Fractions 

Decimals 

Steel 

Iron 

Up  to  76  In.  Wide 

0000000 

1-2 

.5 

20.320 

20.00 

5  per  cent. 

000000 

i6-;« 

.46875 

19.060 

18.75 

00000 

7-1(5 

.4.375 

17.780 

17.50 

6  per  cent. 

0000 

13-32 

.40625 

16.510 

16.25 

000 

8-8 

.:i75 

15.240 

15.00 

7  per  cent. 

00 

ii-;i2 

..34375 

13.970 

13.75 

0 

6-l() 

.3125 

12.700 

12  50 

8  per  cent. 

1 

•►-32 

.28125 

11.4.30 

11.25 

2 

17-44 

.2(55(52 

10.795 

10.(525 

Up  to  50  in. 

3 

1-4 

.25 

10.1(50 

10.00 

wide 

4 

15-1)4 

.23437 

9.52.1 

9.375 

5 

7-32 

.21875 

8.890 

8.75 

7  per  cent. 

6 

VMW 

.20:^12 

8.255 

8.125 

7 

3-16 

.1875 

7.(520 

7.5 

8 

ll-<54 

.17187 

6.i>85 

6.875 

. 

9 

5-32 

.15(525 

6.350 

6.25 

^  8%  per  cent. 

10 

•M)4 

.14062 

5.715 

5.(525 

1 10  per  cent. 

11 

1-8 

.125 

5.080 

5.00 

12 

7-(54 

.10937 

4.445 

4.375 

13 

3-32 

.09374 

3.810 

3.75 

14 

5-64 

.07812 

3.175 

3.125 

15 

9-128 

.07031 

2.857 

2.812 

1(5 

1-16 

.0625 

2.540 

2..'>0 

17 

9-1(50 

.05625 

2.28(5 

2.25 

18 

1-20 

.a5 

2.032 

2. 

\\) 

7-160 

.04375 

1.778 

1.75 

20 

;i-«o 

.0.375 

1.524 

l.-W 

21 

11-.'V20 

.03437 

1.397 

1.375 

22 

1-32 

.03125 

1.270 

1.2.5 

23 

i>-320 

.02812 

1.143 

1.125 

24 

1-40 

.025 

1016 

25 

7-320 

.02187 

1.389 

!875 

26 

3-160 

.01875 

.762 

.75 

27 

\um 

.01718 

.698 

.(587 

28 

1-64 

.01562 

.635 

.623 

29 

9-(i40 

.01406 

.571 

.5(52 

30 

1-80 

.0125 

.508 

.5 

31 

l-iy^O 

.010t)3 

.694 

.437 

.S2 

13-1280 

.01015 

.413 

.40(5 

33 

3-320 

.00937 

.381 

.375 

.      34 

11-1280 

.00859 

..349 

..343 

35 

5-640 

.00781 

.317 

.312 

.36 

9-1280 

.00703 

.28.'> 

.281 

37 

17-2560 

.00f5(ht 

.271 

.2(55 

38 

1-160 

.00625 

.254 

.25 

122 


VOCATIONAL   MATHEMATICS 


Weights  and  Arbas  of  Round,  Square,  and  Hexagon  Steel 
Weight  of  one  cubic  inch  =  .2836  lb.        Weight  of  one  cubic  foot  =  490  lb. 


c« 

Abea 

=  DlAM.«  X 

.7854    1 

Akea  =  Side2  X  1 

Area  =  DiAM.2x. 866 

c - 

Hound 

Sqtiare 

Ilea'agon 

Weight 
Per  Inch 

Area 
Square 
Inches 

Circum- 
ference 
Inches 

Weight 
Per 
Inch 

Area 
Square 

Inches 

Weight 
Per 

Inch 

Area 
Square 
Inches 

1-32 
1-16 
S-S2 
1-8 

.0002 
.0009 
.0020 
.0035 

.0008 
.0031 
.0069 
.0123 

.0981 
.1963 
.2995 
.3927 

.0003 
.0011 
.0025 
.0044 

.0010 
.0039 
.0088 
.0156 

.0002 
.0010 
.0022 
.0038 

.0008 
.0034 
.0076 
.0135 

5-32 
8-16 
7-32 
1-4 

.0054 
.0078 
.0107 
.0139 

.0192 
.0276 
.0376 
.0491 

.4908 
.5890 
.6872 
.7854 

.0069 
.0101 
.0136 
.0177 

.0244 
.0352 
.0479 
.0625 

.0060 
.0086 
.0118 
.0154 

.0211 
.0304 
.0414 
.0540 

9-32 

5-16 

11-32 

3-8 

.0176 
.0218 
.0263 
.0313 

.0621 
.0767 
.0928 
.1104 

.8835 

.9817 

1.0799 

1.1781 

.0224 
.0277 
.0335 
.0405 

.0791 
.0977 
.1182 
.1406 

.0194 
.0240 
.0290 
.0345 

.0686 
.0846 
.1023 
.1218 

13-82 

7-16 

15-82 

1-2 

.0368 
.0426 
.0489 
.0557 

.1296 
.1503 
.1726 
.1963 

1.2762 
1.3744 
1.4726 
1.5708 

.0466 
.0.543 
.0623 
.0709 

.1651 
.1914 
.2197 
.2500 

.0405 
.0470 
.0540 
.0614 

.1428 
.1658 
.1903 
.2161 

17-82 

9-16 

19-32 

5-8 

.0629 
.0705 
.0785 
.0870 

.2217 
.2485 
.2769 
.3068 

1.6689 
1.7671 
1.8653 
1.9635 

.0800 
.0897 
.1036 
.1108 

.2822 
.3164 
.3526 
.3906 

.0693 
.0777 
.0866 
.0959 

.2444 
.2743 
.3053 
.3383 

21-32 

11-16 

23-32 

3-4 

.0959 
.1053 
.1151 
.1253 

.3382 
.3712 
.4057 
.4418 

2.0616 
2.1598 
2.2580 
2.3562 

.1221 
.1340 
.1465 
.1622 

.4307 
.4727 
.5166 
.5625 

.1058 
.1161 
.1270 
.1382 

.3730 
.4093 
.4474 

.4871 

25-32 

13-16 

27-32 

7-8 

.1359 
.1470 
.1586 
.1705 

.4794 
.5185 
.5591 
.6013 

2.4543 
2.5525 
2.6507 
2.7489 

.1732 
.1872 
.2019 
.2171 

.6103 
.6602 
.7119 
.7656 

.1499 

!   .1620 

.1749 

.1880 

.5286 
.5712 
.6165 
6631 

29-32 
15-16 
31-32 

1 

.1829 
.1958 
.2090 
.2227 

.6450 
.6903 
.7371 

.7854 

2.8470 
2.9452 
3.0434 
3.1416 

.2329 
,2492 
.2661 
.2836 

.8213 

.8789 

.9384 

1.0000 

.2015 
.2159 
.2305 
.2456 

.7112 
.7612 
.8127 
.8643 

1  1-16 
1  1-8 
1  3-16 
1  1-4 

.2515 
.2819 
.3141 
.3480 

.8866 

.9940 

1.1075 

1.2272 

3.3379 
3.5343 
3.7306 
3.9270 

.3201 
.3589 
.4142 
.4431 

1.1289 
1.2656 
1.4102 
1.5625 

.2773 
.3109 
.3464 
.3838 

.9776 
1.0973 
1.2212 
1.3531 

1  5-16 
1  3-8 
1  7-16 
1  1-2 

.3837 
.4211 
.4603 
.5012 

1.3530 
1.4849 
1.6230 
1.7671 

4.1233 
4.3197 
4.5160 
4.7124 

.4885 
.5362 
.5860 
.6487 

1.7227 
1.8906 
2.0664 
2.2500 

.4231 
.4643 
.5076 
.5526 

1.4919 
1.6373 
1.7898 
1.9485 

1  9-16 
1  5-8 

1  11-16 
1  3-4 

.5438 
.5882 
.6343 
.6821 

1.9175 
2.0739 
2.2365 
2.4053 

4.9087 
5.1051 
5.3014 

5.4978 

.6930 
.7489 
.8076 
.8685 

2.4414 
2.6406 

2.8477 
3.0625 

.5996 
.6480 
.6994 
.7521 

2.1143 

2.2847 
2.4662 
2.6522 

1  13-16 

1  7-8 

1  15-16 

2 

.7317 
.7831 
.8361 
.8910 

2.5802 
2.7612 
2.9483 
3.1416 

5.6941 
5.8905 
6.0868 
6.2832 

.9316 

.9970 

1.0646 

1.1342 

3.2852 
3.5156 
3.7539 
4.0000 

.8069 
.8635 
.9220 
.9825 

2.8450 
3.0446 
3.2509 
3.4573 

SHEET    AND    ROD    METAL   WORK 


123 


Wkiohts  and  Areas  of  Round,  Square,  and  Hexagon  Stebl.  — Con- 
tinued 

Weight  of  one  cubic  inch  »  .2S8C  lb.        Weight  of  one  cubic  foot  -  490  lb. 


8« 

Area 

=  DiAM.*  X 

.7854 

Area  =  Side*  x  1 

Arka»Diaii.*x  .866 

IE 

Hound                      1 

Square 

,         Ilevagon 

55 

Weight 
Per  Inch 

Are* 
Square 
Indies 

CHrcum- 
ference 
Inches 

Weight 
Per 
Inch 

Area 
Square 
I  iiuhes 

Weight 
Per 
Inch 

Area 
Square 
Inches 

S  1-1« 
S  1-8 

S  S-16 
11-4 

.9475 
1.0058 
1.0658 
1.1276 

3.3410 
3.5466 
3.7583 
3.9761 

6.4795 
6.6759 
6.8722 
7.0686 

1.2064 
1.2806 
1.3570 
1.4357 

4.2539 
4.5156 
4.7852 
5.0625 

1.0448 
1.1091 
1.1753 
1.2434 

3.6840 
3.9106 
4.1440 
4.3892 

S  6-16 
t  S-8 

S  7-16 
%  1-S 

1.1911 
1.2564 
1.3234 
1.3921 

4.2000 
4.4301 
4.6664 
4.9087 

7.2649 
7.4613 
7.6575 
7.8540 

1.5165 
1.6569 
1.6849 
1.7724 

5.3477 
5.6406 
5.9414 
6.2500 

1.3135 
1.3854 
1.4593 
1.5351 

4.6312 
4.8849 
5.1454 
5.4126 

S6-8 
S  S-4 

S  7-8 
S 

1.5348 
1.6845 
1.8411 
2.0046 

5.4119 
5.9396 
6.4918 
7.0636 

8.2467 
8.6394 
9.0321 
9.4248 

1.9541 
2.1446 
2.3441 
2.5548 

6.8906 
7.5625 
8.2656 
9.0000 

1.6924 
1.8574 
2.0304 
2.2105 

5.9674 
6.5493 
7.1590 
7.7941 

S  1-8 
S  1-4 
S  S-8 
Sl-1 

2.1752 
2.3527 
2.5371 
2.7286 

7.6699 
8.2958 
8.9462 
9.6211 

9.8175 
10.2102 
10.6029 
10.9956 

2.7719 
2.9954 
3.2303 
3.4740 

9.7656 
10.5625 
11.3906 
12.2500 

2.3986 
2.5918 
2.7977 
3.0083 

8.4573 

9.1387 

9.8646 

10.6089 

S  8-8 
SS-4 

S  7-8 

4 

2.9269 
3.1323 
3.3446 
3.5638 

10.3206 
11.0447 
11.7932 
12.5664 

11.3883 
11.7810 
12.1737 
12.5664 

3.7265 
3.9880 
4.2582 
4.5374 

13.1407 
14.0625 
15.0156 
16.0000 

3.2275 
3.4539 
3.6880 
3.9298 

11.3798 
12.1785 
13.0035 
13.8292 

4  1-8 

4  1-4 
4  S-8 
4  1-S 

3.7900 
4.0232 
4.2634 
4.5105 

13.3640 
14.1863 
15.0332 
15.9043 

12.9591 
13.3518 
13.7445 
14.1372 

4.8254 
5.1223 
5.4280 
5.7426 

17.0156 
18.0625 
19.1406 
20.2500 

4.1792 
4.4364 
4.7011 
4.9736 

14.7359 
15.6424 
16.5761 
17.5569 

4  6-8 
4  S-4 

4  7-8 

8 

4.7645 
5.0255 
5.2935 
5.5685 

16.8002 
17.7205 
18.6655 
19.6350 

14.5299 
14.9226 
15.3153 
15.7080 

6.0662 
6.6276 
6.7397 
7.0897 

21.3906 
22.5625 
23.7656 
25.0000 

5.2538 
5.5416 
5.8371 
6.1403 

18.5249 
19.5397 
20.5816 
21.6503 

8  1-8 

6  1-4 
6  S-8 
6  1-S 

5.8504 
6.1392 
6.4351 
6.7379 

20.6290 
21.6475 
22.6905 
23.7583 

16.1007 
16.4934 
16.8861 
17.2788 

7.4496 
7.8164 
8.1930 
8.5786 

26.2656 
27.5624 
28.8906 
30.2500 

6.4511 
6.7697 
7.0959 
7.4298 

22.7456 
23.8696 
25.0198 
26.1971 

6  8-8 
6  S-4 
6  7-8 

6 

7.0476 
7.3643 
7.6880 
8.0186 

24.8505 
25.9672 
27.1085 
28.2743 1 

17.6715 
18.0642 
18.4569 
18.8496 

8.9729 
9.3762 
9.7883 
10.2192 

31.6406 
.33.0625 
34.5156 
36.0000 

7.7713 
8.1214 
8.4774 
8.8420 

27.4013 
28.6361 
29.8913 
31.1765 

6  1-4 
6  1-S 
6S-4 

7 

8.7007 

9.4107 

10.1485 

10.9142 

30.6796 
33.1831 
35.7847 
38.4845 

19.6350 
20.4204 
21.2058 
21.9912 

11.0877 
11.9817 
12.9211 
13.8960 

39.0625 
42.2500 
45.5625 
49.0000 

9.5943 
10.3673 
11.1908 
12.0351 

33.8291 
36.5547 
39.4584 
42.4354 

Tl-1 

8 

12.5291 
14.2553 

44.1786 
50.2655 

23.5620 
25.1328 

15.9520 
18.1497 

56.2500 
64.0000 

13.8158 
16.7192 

48.7142 
66.3169 

Pupils  should  practice  the  use  of  tables  in  order  to  obtain  accurate  results  quickly. 
Additional  tables  like  these  may  be  obulned  from  such  standanl  handbooks  as  Kents. 
Multiply  above  weights  by  .998  for  wrought  iron,  .918  for  ca.st  iron.  ^^Y.V^\  for  cast  brass, 
.1209  for  copper,  and  1.1748  for  phos.  bronze. 


124  VOCATIONAL   MATHEMATICS 

EXAMPLES 

By  means  of  the  table  of  weights  and  areas  of  round,  square, 
and  hexagonal  steel,  solve  the  following  problems : 

1.  Find  the  circumference  of  a  steel  bar  (a)  -^^"  in  thickness ; 
(6)  ^"  in  thickness ;  (c)  \l"  in  thickness ;  {d)  111"  in  thick- 
ness; (e)  3|"  in  thickness. 

2.  Find  the  area  of  a  steel  bar  (a)  |^"  in  diameter ;  (6)  1^^" 
in  diameter;  (c)  3|"  in  diameter;  (d)  4^'  in  diameter;  (e)  5 J" 
in  diameter. 

3.  Find  the  weight  per  inch  of  a  steel  bar  (a)  ||''  in 
diameter;  (h)  ly^' in  diameter ;  (c)  2^^"  in  diameter. 

4.  Find  the  area  of  a  square  bar  (a)  f"  per  side ;  (h)  If" 
per  side ;  (c)  5^  per  side ;  (d)  3|''  per  side. 

5.  Find  the  weight  per  inch  of  a  square  bar  (a)  j^'  in 
thickness ;  (6)  f|"  in  thickness ;  (c)  Iff"  in  thickness. 

6.  Find  the  area  of  a  hexagonal  bar  (a)  i|"  in  thickness ; 
(6)  1\^"  in  thickness ;  (c)  3J"  in  thickness. 

7.  Find  the  weight  per  in.  of  a  hexagonal  bar  (a)  |f"  in 
thickness ;  (6)  |^''  in  thickness ;  (c)  2^^  in  thickness. 

8.  Find  the  weight  of  a  round  bar  of  steel  (a)  7'  long  -f^"_ 
in  diameter ;  (&)  11'  long  lyV  in  diameter ;  (c)  16'  long  2^%" 
in  diameter ;  {d)  IV  long  2J"  in  diameter ;  (e)  13'  long  4|"  in 
diameter. 

9.  Find  the  weight  of  a  round  bar  of  steel  (a)  8'  long  2|" 
in  diameter;  (h)  14'  long  1|"  in  diameter;  (c)  11'  long  lif"  in 
diameter. 

10.  Find  the  weight  of  a  hexagonal  bar  of  steel  (a)  8'  long 
3f"  in  diameter;  {h)  T  long  2J"  in  diameter;  (c)  9'  long  Ifi" 
in  diameter. 

11.  What  is  the  weight  of  a  square  bar  of  wrought  iron  16' 
long  4^"  in  thickness  ? 


SHEET    AND    ROD    METAL   WORK  125 

12.  What  is  the  weight  of  a  hexagonal  bar  of  wrought  iron 
18'  long  2yV'  i"  diameter  ? 

13.  What  is  the  weight  of  a  round  bar  of  cast  iron  14'  long 
Si"  in  diameter? 

14.  What  is  the  weight  of  a  square  bar  of  cast  iron  13'  long 
1|"  in  thickness  ? 

15.  What  is  the  weight  of  a  square  bar  of  cast  brass  15'  long 
l|f "  in  thickness  ? 

16.  What  is  the  weight  of  a  hexagonal  bar  of  cast  brass  8' 
long  If"  in  diameter  ? 

17.  What  is  the  weight  of  a  hexagonal  bar  of  copper  7'  long 
2yV'  in  diameter  ? 

la   What  is  the  weight  of  a  round  bar  of  copper  6'  long  l^y 
in  diameter? 


PART   IV— BOLTS,   SCREWS,   AND   RIVETS 

CHAPTER   VIII 

BOLTS 

The  most  common  forms  of  fastenings  are  bolts,  screws, 
rivets,  pins,  and  nails.  These  are  turned  out  in  large  numbers, 
usually  by  feeding  long  lengths  of  iron  or  steel  rod  into  auto- 
matic machines.  In  order  to  make  these  fasteners  of  the  right 
size  one  has  to  be  familiar  with  the  problems  that  are  con- 
nected with  them. 

The  common  bolt  is  made  in  many  different  sizes  and  is 
usually  held  in  place  by  a  nut  screwed  on  the  end.  There  are 
many  different  kinds  of  bolts  for  the  different  uses  to  which 
they  are  put. 

Rough  Bolts 

The  small  diameter  of  a  rough  bolt  head,  that  is,  the  dis-. 
tance  across  the  flats,  is  1^  times  the  diameter  of  the  bolt,  plus 
1  inch ;  or,  it  may  be  stated  as  follows :   The  diameter  of  a 
rough  bolt  head  =  li^Z)  plus  ^  inch,  D  being  diameter  of  the 
bolt. 

Sometimes  bolts  or  nuts  are  made  from  round  stock  and  cut 
either  square  or  hexagon.  In  such  cases  it  is  necessary  to  hud 
the  proper  diameter  to  which  the  stock  must 
be  turned  in  order  that  it  may  be  milled  to 
size.  Let  A  represent  the  distance  across  the 
flats  on  the  head  of  a  flat  bolt,  and  B  the 
diameter  of  the  round  stock  required  to  make 
a  square-head  bolt.  Square  Head 

126 


BOLTS,   SCREWS,  AND   RIVETS 


127 


Wlieii  the  dimension  A  is  given  and  it  is  necessary  to  turn  a 
piece  that  will  mill  down  to  this  size  and  leave  full  corners, 
multiply  A  by  1.414.     The  product  is  the  desired  size  B, 


B  =  y/WTA^  Why  ? 

B  =  y/¥A^  =  AVi  =  1.414  A 

When  ^1,  the  distance  across  the  flats,  is  given,  and  it  is 
necessary  to  turn  a  piece  that  will  mill  down  to  this  size  and 
leave  full  corners,  multiply  A  by  1.155.  The  product  is  the 
desired  size  {B)  of  the  hexagon  head. 


Let 


X 

2 


(ir =(fr-(fy  -^^ 


4  ■■  16       4 
4  52  =  //-'  -f  4  X2 
3  J?^  =  4  X2 

4 


B^ 


3 


X2 


B=2X^2_XV3=1.165X 
V3  3 


Hexagonal  Head 


Example, — To  what  diameter  should  a  piece  of  stock  be 
turned  in  order  that  it  may  have  full  corners  when  milled 
down  six-sided  to  \\"  across  the  flats? 

li"  X  1.155  =  1.7325",  diameter  of  the  blank.     Ans. 


EXAMPLES 

1.  What  is  the  size  of  stock  required  to  make  the  following 
bolts  having  hexagonal  heads  ?  (a)  Body  .}"  diam.  (6)  Body 
J"  diam.     (c)  Body  I"  diam.     (d)  Body  V  diam. 

2.  What  is  the  size  of  stock  required  to  make  the  following 
bolts  with  square  heads?  (a)  Body  \"  diam.  (6)  Body  \" 
diam.      (c)  Body  |"  diam.     {d)  Body  \"  diam. 


128 


VOCATIONAL   MATHEMATICS 


EXAMPLES 

1.  To  what  diameter  should  a  piece  of  stock  be  turned  so 
that  it  ma}^  have  full  corners  when  milled  down  square  to  1|" 
across  the  fiats  ? 

2.  To  what  diameter  should  a  piece  of  stock  be  turned  in 
order  that  it  may  have  full  corners  when  milled  down  six-sided 
to  1|"  across  the  flats  ? 

Sizes  of  Standard  Hexagon  Head  Bolts 


Size  of 

DiAM.    OK 

Bolt 

Thickness 
OF  Heads 

Hexagon  or 

UlSTANCE 

Across 
Corners 

Threads 

Tap  Drill 

In. 

In. 

Across  the 

Flats 

In. 

In. 

Per  Inch 

In. 

i 

i 

1% 

20 

t\ 

t\ 

H 

u 

n 

18 

C 

M 

H 

¥i 

16 

N 

tV 

M 

11 

If 

14 

S 

tV 

i 

1 

13 

if 

T? 

n 

M 

V. 

12 

If 

H 

ii*. 

1/^ 

11 

II 

f 

n 

i-A 

10 

f 

II 

IrV 

m 

9 

H 

if 

n 

ii 

8 

II 

M 

m 

2/2 

7 

fi 

1 

2 

h% 

7 

1/t 

h\ 

2A 

n 

0 

Ui 

ifV 

2| 

2| 

6    . 

m 

1/3. 

2/. 

2-it 

5^ 

111 

If 

2| 

•3A 

5 

H 

HI 

m 

3M 

5 

If 

2 

1^ 

H 

3t 

H 

111 

^ 

H 

n 

4t^. 

41 

1.962  =  If  ^ 

^ 

iH 

H 

4i 

4 

2.176  =  2/^ 

2f 

2i 

H 

4|f 

4 

2.426  =  2j.V 

3 

2A 

^ 

H 

3^ 

2.629  =  2f  ^ 

Notice  that  size  of  hexagon  is  equal  to  diameter  of  bolt  +  \  diameter  of 
bolt  +  I  of  an  inch,  and  also  that  thickness  of  head  is  \  of  hexagon  in 
every  case.     The  thickness  of  nut  is  equal  to  the  diameter  of  bolt. 


BOLTS.   SCREWS,   AKD   UIVKTS 


129 


3.  To  what  diameter  should  a  piece  of  stock  be  turned  so 
that  it  may  have  full  corners  when  milled  down  sc^uare  to  2J" 
across  the  flats  ? 

4.  To  what  diameter  should  a  piece  of  stock  be  turned  so 
that  it  may  have  full  corners  when  milled  down  six-sided  to 
2 J"  across  the  flats  ? 

Size  across  Corners  of  Squares 


SiXK  OF  SqUARK,  In. 

DiAUONAL 

Size  of  Sqitabr,  In. 

Diagonal 

} 

.177 

1.4141 

^. 

.205 

1.590      . 

\ 

.354 

1.768 

^  . 

.442 

1.945 

i 

.530 

2.121 

A 

.619 

2.298 

i 

.707 

2.476 

A 

.706 

2.652 

f 

.884 

2 

2.828 

n 

.972 

^ 

3.005 

i 

1.061 

2\ 

3.182 

H 

1.149 

^ 

3.535 

1 

1.237 

2f 

3.889 

H 

1.326 

3 

4.243 

EXAMPLES 
Use  the  table  to  obtain  the  size  of  bolts. 

1.  Find  the  distance  across  the  corners  of  a  hexagon  head 
bolt  (a)  with  a  diameter  of  1  J" ;  (6)  with  a  diameter  of  2J" ; 
(c)  with  a  diameter  of  1\". 

2.  Find  the  diagonal  distance  across  a  square  bolt  with  size 
ofsquare(a)-fV';  (/>)  A"- 

3.  (a)  If  a  bolt^heading  machine  has  the  following  daily 
output,  23:^0,  2060,  1950,  2420,  2310,  2030,  what  is  the  average 


130  VOCATIONAL  MATHEMATICS 

daily  output  ?     (6)  What  would  be  the  daily  wage  at  13  cents 
per  100  bolts  ?     (c)  The  weekly  wage  ? 

4.  A  blacksmith  requires  six  pieces  of  steel  of  the  following 
lengths:  If",  2|",  2j\'\  ^\\",  1\^",  2".  How  long  a  piece  of 
steel  will  be  necessary  to  make  them,  if  ^'  is  allowed  for  each 
in  finishing  ? 

5.  A  machinist  has  to  make  five  bolts  from  the  same  size 
bar.  One  bolt  is  to  be  1-^'  over  all,  another  2^",  another  2-^%", 
another  3//',  and  the  last  2f"..  How  much  stock  will  he  need 
if  he  allows  ^"  for  each  cut-off  ? 

Rivets 

One  of  the  simplest  and  most  efficient  metal  fastenings  which 
has  been  extensively  used  is  the  rivet.  It  resembles  the  bolt, 
but  it  can  be  removed  only  by  chipping  off  the  head,  while  the 
bolt  can  be  taken  off  by  removing  the  nut.  Rivets,  like  bolts 
and  nails,  are  quickly  turned  out  by  the  thousand  with  the 
aid  of  automatic  machines. 

EXAMPLES 

1.  What  must  be  the  length  of  a  bolt  under  the  head,  to  go 
through  9^y  thickness  of  plank  and  allow  IJ"  outside  for  tak- 
ing a  nut  ? 

2.  Whgit  must  be  the  length  of  a  bolt  under  the  head,  to  go 
through  7^\"  thickness  of  plank  and  allow  If"  outside  for  tak- 
ing a  nut  ? 

3.  (a)  How  many  rivets  can  be  made  in  a  bolt  machine, 
from  a  round  iron  rod  6'  long,  if  each  rivet  requires  2 J"  of  bar  ? 
(6)  If  the  bar  weighs  |  lb.  per  ft.,  how  many  rivets  weighing 
\  lb.  apiece  can  be  made  from  it  ?  (c)  How  much  waste  will 
there  be  in  each  case  ? 

4.  What  is  the  total  thickness  of  three  plates  riveted  to- 
gether, each  plate  i^"  thick  ?  What  would  be  the  total  thick- 
ness of  five  such  plates  ? 


BOLTS,   SCREWS,  AND  RIVETS  131 

5.  What  is  the  thickness  of  a  steel  plate  that  is  only  |  the 
thickness  of  a  -j^"  plate  ? 

6.  A  blacksmith  and  his  helper  made  192  bolt  dogs  in  18J^ 
hours.  They  received  6.}  cents  apiece  for  them,  (a)  How  much 
did  they  both  receive  per  hour?  (b)  If  the  blacksmith  re- 
ceived G65  %  of  the  money,  how  much  did  each  receive  per 
hour? 

7.  How  many  feet  of  round  iron  weighing  2.67  lb.  per  foot 
will  be  required  to  make  87  rivets  weighing  2^  lb.  apiece,  not 
counting  waste  ?     Give  answer  in  feet  and  decimals  of  foot. 

8.  How  many  rivets  weighing  7.J  ounces  each  can  be  made 
from  15'  of  round  iron  weighing  1.5  lb.  per  foot  ?  How  much 
waste  will  there  be  ?  , 

9.  If  the  drawing  of  an  armor  bolt  is  made  J  size  and  the 
length  of  the  drawing  measures  7J",  what  will  it  measure  if 
made  to  the  scale  of  3''  =  1'  ?  What  is  the  actual  length  of  the 
bolt? 

10.  If  the  drawing  of  an  armor  bolt  is  made  \  size  and  the 
length  of  the  drawing  measures  9|",  what  will  it  measure  if 
made  to  the  scale  of  4"  =  1'?  What  is  the  actual  length  of 
the  bolt? 

11.  The  over-all  length  of  a  threaded  bolt  is  7f",  the  thick- 
ness of  the  head  is  I",  and  the  other  end  of  the  bolt  is  threaded 
for  a  distance  of  2^" ;  what  is  the  length  of  the  shank  between 
the  under  side  of  the  head  and  the  threaded  part  of  the  bolt  ? 

Nails 

W^ooden  objects  are  often  held  together  by  means  of  nails. 
There  are  two  kinds  of  nails:  cut  and  wire.  The  wire  nails 
are  more  commonly  used,  as  they  penetrate  the  wood  without 
splitting  it  as  the  cut  nails  do.  They  have  different  kinds  of 
heads,  according  to  the  use  for  which  they  are  intended. 


132 


VOCATIONAL  MATHEMATICS 


The  origin  of  the  common  terms  "  sixpenny,"  "  tenpenny,"  etc.,  as 
applied  to  nails,  though  not  generally  known,  is  involved  in  no  mystery. 
Nails  have  been  made  a  certain  number  of  pounds  to  the  thousand  for 
many  years  and  are  still  reckoned  in  that  way  in  England,  a  tenpenny 
behig  a  thousand  nails  to  ten  pounds,  a  sixpenny  a  thousand  nails  to  six 
pounds,  a  twentypenny  weighing  twenty  pounds  to  the  thousand ;  and  in 
ordering  buyers  call  for  the  three-pound,  six-pound,  or  ten-pound  variety, 
etc.,  until,  by  the  Englishmen's  abbreviation  of  "pun"  for  "pound," 
the  abbreviation  has  been  made  to  stand  for  penny,  instead  of  pound,  as 
originally  intended. 


Length  and  Number  of  Cut  Nails  to  the  Pound 


SiZK 

in 

0 

d 

0 
0 

1 

K 

■< 

S 
■< 
0 

s 
^ 

M 

< 

H 
0 

f 

|in. 

800 

f 

lin. 

500 

2d 

1  in. 

800 

1100 

1000 

376 

3d 

li  in. 

480 

720 

760 

224 

4d 

11  in. 

288 

623 

368 

180 

398 

5d 

If  in. 

200 

410 

130 

6d 

2  in. 

1H8 

95 

84 

268 

224 

126 

96 

7d 

2^  in. 

124 

74 

64 

188 

98 

82 

8d 

21  in. 

88 

62 

48 

146 

128 

75 

68 

9d 

2|  in. 

70 

53 

36 

130 

110 

65 

lOd 

3  in. 

58 

46 

30 

102 

91 

55 

28 

12d 

3Jin. 

44 

42 

24 

76 

71 

40 

16d 

31  in. 

35 

38 

20 

62 

54 

27 

22 

20d 

4  in. 

23 

33 

16 

54 

40 

14^ 

30d 

4^  in. 

18 

20 

33 

m 

40d 

5  in. 

14 

27 

^ 

60d 

6^  in. 

10 

8 

60d 

6  in. 
6^  in. 

7  in. 

8  in. 

8 

6 

BOLTS,   SCREWS,   AND   RIVETS 


133 


EXAMPLES 

1.  How  many  Jd  nails  are  there  in  3  pounds  ? 

2.  How  many  Jd  nails  are  there  in  4  pounds  ? 

3.  How  many  2d  nails  are  there  in  2  pounds  ? 

4.  How  many  7d  nails  are  there  in  8  pounds  ? 

5.  How  many  16d  nails  are  there  in  1(>  pounds  ? 

6.  How  long  is  (a)  an  8d  nail?  (b)  a  40d?  (o)  a  16d? 
(d)  a  9d  ? 

7.  How  long  is  (a)  a  7d  nail?  (6)  a  5d  ?  (c)  a  12d  ? 

Tacks 

Tacks  are  used  to  fasten  thin  pieces  of  material  to  wood. 
They  vary  in  form  and  size.  The  size  is  represented  by  a 
number,  as  16  oz.  tacks,  24  oz.  tacks ;  a  No.  1  tack  is  called  a 
one-ounce  tack. 

Ndmbbr  of  Tacks  in  a  Pound 


Title 

Length 

No.  PER  Lb. 

Title 

Lexotii 

N(».  PER  Lll. 

1    ounce 

A  inch 

10,000 

10  ounce 

H    inch 

1,600 

IJ  ounce 

y'j  inch 

10,606 

12  ounce 

1     inch 

1,332 

2    ounce 

i    inch 

8,000 

14  ounce 

\i    inch 

1,143 

2\  ounce 

A  i"^^ 

6,400 

10  ounce 

1      inch 

1,000 

3    ounce 

f    inch 

5,3.32 

18  ounce 

U    inch 

888 

4    ounce 

^  inch 

4,000 

20  ounce 

1      inch 

800 

6    ounce 

A  inch 

2,060 

22  ounce 

ly^g  inch 

727 

8    ounce 

f    inch 

2,000 

24  ounce 

1\    inch 

666 

EXAMPLES 

1.  How  many  tacks  are  there  in  8  lb.  of  1  oz.  or  No.  1  tacks  ? 

2.  How  many  tacks  are  there  in  13  lb.  of  2^  oz.  or  No.  2^ 
tacks? 

3.  How  many  tacks  are  there  in  7  lb.  of  8  oz.  or  No.  8  tacks  ? 


134  VOCATIONAL  MATHEMATICS 

4.  How  many  tacks  are  there  in  6  lb.  of  12  oz.  or  No.  12 
tacks  ? 

5.  How  many  tacks  are  there  in  17  lb.  of  20  oz.  or  No.  20 
tacks  ? 

Screws 

Pieces  of  wood  and  of  metal  are  often  fastened  together  by 
means  of  screws  instead  of  by  nails,  especially  if  it  is  desirable 
to  separate  the  parts  at  any  time.  Screws  are  made  of  iron, 
steel,  or  brass  and  have  either  a  flat,  fillister,  or  round  head. 
When  it  is  desired  to  have  the  head  of  the  screw  flush  with  the 
surface,  the  flat  head  type  is  used.  Screws  are  made  on  auto- 
matic screw  machines  into  which  wire  of  various  sizes  may  be 
fed.  The  machines  are  so  constructed  that  they  turn  out 
large  numbers  of  screv/s  all  complete  in  a  very  short  time. 

iMmmmiQ^  UiUMlliillig!^  iuiiimiiiiiirfi 

^^^^^^MHH|     ^Wfffff'^^B^^     ^^^^^^^^^^jg 

flat  head  round  head  fillister  head 

Iron  Machine  Screws 

Screw  threads  are  divided  into  two  classes :  first,  those  used 
for  fastening ;  and  second,  those  used  in  large  machines  for 
communicating  motion.  The  screw  threads  used  for  communi- 
cating motion  with  which  the  mechanic  has  to  deal  are  pro- 
duced by  a  cutting  process  in  which  the  thread  is  formed  from 
the  solid  piece  of  stock  by  means  of  a  single  pointed  cutting 
tool  in  a  lathe.  Screws  used  for  fastening  are  made  by  means 
of  taps  and  dies.  The  tap  is  a  tool  used  to  produce  internal 
threads,  and  the  die  is  a  tool  used  to  cut  the  external  threads. 
The  screw  thread  is  applied  in  many  ways,  but  the  most  com- 
mon use  is  that  of  fastening  together  the  various  parts  of 
machines,  etc. 

We  find  the  mechanic  using  many  different  forms  of  bolts 
and  screws   to  meet  the  needs  of   industries.      In   order  to 


BOLTS.  SCREWS.  AND   RIVETS 


135 


specify  a  particular  f?rade  of  bolt  or  screw  it  is  necessary  to 
mention  (a)  shape  or  form  of  head,  (b)  pitch  or  number  of 
threads  to  the  inch,  (c)  shape  of  thread,  (rf)  outline  of  body, 
barrel  or  stem,  (e)  size  of  diameter,  (/)  direction  of  thread,  as 
right  or  left  hand,  (g)  length,  (/i)  material,  as  brass,  iron,  etc. 
There  are  four  different-shaped  threads  in  common  use  in 
the  United  States :  1.  The  V  thread ;  2.  The  U.  S.  standard ; 
3.  The  Acme  standard  or  worm  ;  4.  The  square  thread. 


Sharp  V  Thread 

*PITCH- 


U.  S.  Standard  Thread 


AcMR  Standard  or  Worm  Thread 


TTTTZy 


P/TCH 


Squabb  Thread 
1  Sometimes  called  modified  square  thread. 


136 


VOCATIONAL  MATHEMATICS 


The  different  screws  are  formed  by  cutting  a  spiral  groove 
around  a  cylinder.  The  projecting  stock  between  the  grooves 
is  called  the  land  or  thread.  There  may  be  any  number  of 
threads ;  to  every  groove  there  is  an  accompanying  thread. 


RiGHT-ILvxD,  Single  V-Thread,  8  Threads  to  the  Inch 


Nut 


A  screw  that  has  one  thread  is  called  single-threaded ;  one 
having  two  threads,  double-threaded;  three  threads, 
triple-threaded;  etc.  The  cutting  of  the  spiral 
groove  or  grooves  is  called  cutting  or  threading  a 
screw.  A  nut  is  a  piece  of  iron  or  steel  with  a 
threaded  hole  which  goes  over  a  screw,  and  will 
turn  off  and  on  the  screw. 

Lead  of  a  Screw.  —  The  distance  that  a  thread  advances  in 
one  turn  is  called  the  lead  of  the  screw.  In  a  single-threaded 
screw  the  lead  is  equal  to  the  distance  occupied  by  one  thread  ; 
and  when  the  nut  has  made  one  complete  turn,  it  has  advanced 
one  thread  upon  the  screw.  In  a  double-threaded  screw  the 
nut  advances  two  threads  with  each  complete  turn  ;  when  the 
lead  is  three  threads,  the  nut  advances  three  threads  in  one 
turn.  In  general,  the  lead  can  be  divided  by  any  number  of 
threads,  the  advance  of  any  one  of  these  threads  in  one  turn 
being  always  equal  to  the  lead. 

One  complete  revolution  of  a  single-threaded  screw  or  the 
lead  of  a  screw,  if  the  screw  had  twelve  threads  to  the  inch, 
would  be  one  twelfth  of  an  inch. 


BOLTS,   SCREWS,  AND   RIVETS  137 

Threads  to  an  Inch.  —  By  placing  a  scale  upon  a  screw  as  in 
the  figures  on  page  130,  the  number  of  thread-windings  or  coils 
in  an  inch  can  be  counted.  These  windings  or  coils  are  called 
threads,  and  the  number  of  coils  to  an  inch  is  called  the  number 
of  threads  to  an  indi.  The  thread  commences  at  the  root  or 
bottom  of  the  screw.  To  measure  screws  for  the  number  of 
threads  per  inch,  the  measurement  must  begin  at  the  point 
of  the  thread.  Place  the  end  of  the  scale  in  line  with  this 
portion  and  then  count  the  number  of  threads  within  the  one- 
inch  line. 

Pitch.  —  T?ie  distance  from  the  center  of  one  thread  to  the  cen- 
ter of  the  next  thread,  measured  in  a  line  parallel  to  the  axis,  is 
tJie  pitch  of  the  thread,  or  the  thread-pitch.  Divide  1"  by  the 
number  of  threads  to  1"  and  the  quotient  is  the  thread-pitch. 
The  threads  to  1"  and  the  thread-pitch  are  reciprocals  of  each 
other. 

In  a  single-threaded  screw  the  pitch  is  equal  to  the  lead.  In 
a  double-threaded  screw  the  pitch  is  half  the  lead;  in  a  triple- 
threaded  screw  the  pitch  is  \  the  lead;  and  so  on.  When  the 
thread  inclines  so  as  to  be  nearer  the  right  hand  at  the  under 
side,  or  clockwise,  it  is  a  right-hand  thread.  When  the  under 
side  is  toward  the  left,  or  counter  clockwise,  the  thread  is  left- 
handed.  Again,  when  a  right-hand  screw  turns  in  a  direction 
to  move  its  upper  side  away  from  the  eye,  the  thread  appears 
to  move  toward  the  right;  while  a  left-hand  thread  moves 
toward  the  left. 

The  term  turns  to  an  inch  means  the  number  of  times  a 
screw  must  be  turned  around  to  advance  one  inch.  If  a  screw 
makes  four  turns  in  advancing  one  ioch,  its  lead  is  \,  and  it 
has  4  turns  to  the  inch.  Divide  one  inch  by  the  lead,  and  the 
quotient  is  the  number  of  turns  that  the  screw  makes  in  ad- 
vancing one  inch.  If  a  screw  does  not  advance  exactly  an  inch 
in  a  whole  number  of  turns,  or  if  it  does  not  advance  some 
whole  number  of  inches  in  one  turn,  it  is  said  to  have  a  frac- 
tional thread.     In  any  screw  divide  any  number  of  turns  by 


138  VOCATIONAL  MATHEMATICS 

the  number  of  inches  occupied  by  these  turns  and  the  quotient 
will  be  the  turns  to  an  inch.  Thus,  a  screw  that  turns  96 
times  in  12.005  inches,  turns  96  -^  12.005,  or  7.9967  turns  in 
one  inch. 

EXAMPLES 

1.  How  many  turns  to  the  inch  has  (a)  a  screw  that  ad- 
vances 25  inches  in  200  turns  ?  (b)  a  single-threaded  screw 
of  9  threads  to  the  inch  ?  (c)  a  double-threaded  screw  with 
8  threads  to  the  inch  ?  (d)  a  'double-threaded  screw  with  6 
threads  to  the  inch  ?  (e)  a  single-threaded  screw  with  24 
threads  to  the  inch  ? 

2.  AVhat  is  the  lead  of  a  single-threaded  screw  if  it  has 
(a)  five  threads  to  the  inch  ?  (6)  eight  threads  to  the  inch  ? 
(c)  fifteen  threads  to  the  inch  ?  (d)  twenty-eight  threads  to 
the  inch  ? 

3.  What  is  the  lead  of  a  double-threaded  screw  if  it  has 
eight  threads  to  the  inch  ? 

4.  What  is  the  lead  of  a  single-threaded  screw  if  it  has 
twenty-two  threads  to  the  inch  ? 

5.  What  is  the  lead  of  a  single-threaded  screw  if  it  has 
twenty-four  threads  to  the  inch  ? 

6.  A  jack  screw  has  three  threads  to  the  inch ;  how  far  does 
it  move  in  ^  of  a  revolution  ? 

7.  A  jackscrew  has  three  threads  to  the  inch  ;  how  far  does 
it  move  in  i  of  a  revolution  ? 

8.  What  is  the  thread  pitch  of  a  single-threaded  screw  that 
has  18  threads  to  the  inch  ? 

9.  What  is  the  thread  pitch  of  a  double-threaded  screw 
that  has  8  threads  to  the  inch  ? 

10.  Let  each  pupil  have  five  different  kinds  of  screws,  and 
tell  the  number  of  threads  to  the  inch  of  each  screw. 


BOLTS,   SCREWS,   AND   RIVETS  139 

The  Micrometer 

Accurate  mathematical  work  in  measuring  diameter  is  done 
with  the  micrometer  caliper.  With  this  instrument  thousandths 
of  an  inch  may  easily  be  found.  The  micrometer  is  easily 
adjusted,  finely  graduated,  and  has  stamped  on  its  yoke  the 
fractions  and  decimal  equivalents  which  may  be  needed  in 
close  measuring.  The  parts  of  the  micrometer  best  known  are 
the  screw  J  the  linhj  the  thimble. 


MiCKOMKTER  CALIPER 

The  screw  of  the  micrometer  is  covered  by  the  thimble  to 
protect  it  from  dust  and  wear.  By  turning  the  thimble  we 
move  the  screw  back  and  forward,  increasing  or  decreasing 
the  distance  between  the  measuring  points  of  the  micrometer 
and  so  opening  or  closing  the  instrument  for  larger  or  smaller 
diameters.  One  complete  revolution  of  the  thimble  changes 
the  opening  of  the  caliper  .025,  and  as  the  pitch  of  the  screw 
in  the  caliper  is  40  per  inch  and  the  circumference  of  the 
thimble  graduated  into  25ths,  the  turn  from  one  of  these  to 
the  next  makes  the  caliper  opening  .001. 

The  heel  is  graduated  in  a  straight  line  parallel  with  the 
screw  length  and  conforms  to  the  pitch  of  the  screw,  each 
division  being  .025  inch,  and  the  fourth  division,  which  is 
.100,  is  made  on  the  frame  with  the  figure  1,  the  eighth,  with 
2,  etc.    When  the  thimble  is  turned  one  complete  revolution, 


140  VOCATIONAL  MATHEMATICS 

the  screw  advances  one  fortieth  of  an  inch  and  one  twenty-fifth 
of  one  fortieth  is  .001.  In  using  the  micrometer  care  must  be 
taken  to  get  the  proper  touch  with  the  instrument  or  it  may 
be  crowded  over  with  an  error  of  one  half  thousandth  more 
or  less  than  the  actual  size  of  the  work  required.  The  microm- 
eter is  a  very  delicate  instrument  and  must  be  kept  away  from 
excessive  heat  or  cold  as  expansion  or  contraction  of  the  metal 
will  cause  it  to  become  inaccurate.  In  close  work  the  heat  of 
the  hand,  when  it  is  held  too  long,  will  change  a  micrometer 
reading. 

EXAMPLES 

1.  If  the  screw  of  a  micrometer  has  40  threads  to  the  inch, 
how  far  will  it  move  in  (a)  one  complete  revolution ;  (b)  ^  rev- 
olution ;  (c)  \  revolution ;  (cZ)  f  revolution ;  (e)  J^  of  a  revo- 
lution ? 

2.  What  is.  the  lead  of  the  screw  in  the  above  micrometer  ? 

3.  If  the  screw  of  a  micrometer  has  60  threads  to  the  inch, 
what  is  the  lead  ? 

4.  In  example  3,  how  far  will  it  advance  in  (a)  ^  revolu- 
tion ;  (b)  \  revolution ;  (c)  f  revolution  ? 

V-Shaped  Thread 

The  common  V-shaped  thread  is  a  thread  having  its  sides  at 
an  angle  of  60  degrees  to  each  other,  and  perfectly  sharp,  top 
and  bottom.  The  objections  to  using  this  thread  are  that  the^ 
top  is  so  sharp  that  it  is  injured  by  the  slightest  accident  and 
in  using  the  taps  and  dies  in  making  it  the  fine,  sharp  edge  is 
quickly  lost,  causing  constant  variation  in  fitting. 

Formula: 

P=  Pitch  = 


No.  of  threads  per  inch 
JD  =  Depth  =  P  X  .8660 


BOLTS,  SCREWS,  AND   RIVETS  141 

The  diameter  of  the  root  (effective  diameter)  of  the  thread 
is  found  by  multiplying  the  product  above  by  2,  and  then  sub- 
tracting this  double  depth  from  the  diameter  of  the  screw. 

A  formula  is  used  to  tind  the  size  of  a  tap  drill  to  use  in 
connection  with  a  tap  of  a  given  size  which  has  a  given  number 
of  threads  per  inch,  as : 

Let  7=  diameter  of  the  tap,  or  size  of  the  thread  the  nut 
is  to  tit; 
N=  number  of  threads  per  inch  ; 
S  =  size  at  root  of  the  thread,  or  size  of  the  tap  drill. 

N 
Example.  —  What  must  the  size  be  of  a  tap  drill  for  a  1-inch 
V-thread  tap  or  1-inch  bolt  having  8  threads  per  inch  ? 
According  to  the  formula : 

^  =  1  _  hl^  OT  S  =  l-  .216  or  S  =  .784  inch,  Ans. 
8 

By  referring  to  the  Table  of  Decimal  Equivalents  ^  the  drill 
nearest  in  size  to  .784  is  ^  inch,  which  will  cut  a  trifle  larger 
and  therefore  will  be  right. 

United  States  Standard  Thread 

The  United  States  Standard  Thread  has  its  sides  also  at  an 
angle  of  60  degrees  to  each  other,  but  the  top  is  cut  off  to  the 
extent  of  one  eighth  of  its  pitch  and  the  same  quantity  filled 
in  at  the  bottom.  The  advantages  claimed  for  this  thread  are 
that  it  is  not  so  easily  injured,  that  the  taps  and  dies  retain 
their  size  longer,  and  that  the  bolts  and  screws  made  with 
this  thread  are  stronger  and  have  a  better  appearance.  This 
system  has  been  recommended  by  the  Franklin  Institute  of 
Philadelphia  and  it  is  often  called  the  Franklin  Institute 
Standard.     Although   the  V-shaped  thread  is  the  strongest 

^  See  Appendix  for  Table  of  Decimal  Equivalents. 


142  VOCATIONAL  MATHEMATICS 

form  of  screw  thread,  yet  as  the  thrust  between  the  screw 
and  the  nut  is  parallel  to  the  axis  of  the  screw,  there  is  a 
tendency  to  burst  the  nut.  So  this  form  of  thread  is  unsuitable 
for  transmitting  power. 

As  \  of  the  height  of  the  U.  S.  standard  thread  is  taken  from 
the  top  and  ^  from  the  bottom,  the  thread  is  only  J  as  deep  as 
the  V-form,  so  in  the  formula  for  finding  the  diameter  at  the 
root  of  the  thread  we  use  a  numerator  which  is  but  f  of  the 
numerator  used  for  the  V-thread :  f  of  1.733  is  1.3.  So 
the  formula  is 

Example.  —  What  should  the  size  of  a  tap  drill  be  for  a  one- 
inch  U.  S.  S.  tap? 

As  the  U.  S.  standard  is  not  only  a  thread  of  a  certain  form 
but  also  of  a  given  pitch  for  each  diameter  of  screw,  by  re- 
ferring to  the  table  of  United  States  Standard  Screw  Threads 
we  find  that  one-inch  screws  have  eight  threads  to  the  inch. 

S  =  l-—  ov  S  =  l-  .1625  ovS=  .8375 
8 

In  the  Table  of  Decimal  Equivalents  .8375  has  no  common 
fraction  equivalent  among  the  sizes  given,  but  as  f f  is  only 
.006  larger  we  would  select  a  drill  of  that  size. 

Formula :      P  =  Pitch  = ^ — - 

No.  of  threads  per  inch 

Z>=Depth  =  Px  .6495 
i^=Flat  =  - 


EXAMPLES 

Solve  the  following  examples  by  the  use  of  the  table. 
1.    How  many  threads  are  there  per  inch  of  the  U.  S.  S.  screw 
with  (a)  I"  diameter  ?    (b)  f "  diameter  ? 


BOLTS,   SCREWS,  AND   RIVETS 


143 


2.  What  is  the  diameter  of  a  screw  with  the  U.  S.  S.  thread 
with  (a)  5  threads  per  inch  ?  (6)  4^  threads  per  inch  ?  (c)  13 
threads  per  inch  ? 

Table  of  United  States  Standard  Screw  Threads 


DiAM.   or   SCRKW 

TiiRKAne  I'KB  Incu 

DiAM.   OK  SCRBW 

Threads  i'kb  Ikcii 

i    in. 

20 

l}in. 

6 

A  in. 

18 

n  in. 

6 

i    in. 

16 

If  in. 

6i 

Ai°- 

14 

If  in. 

5 

J    in. 

13 

l|in. 

6 

A  in. 

12 

2    in. 

^ 

f    in. 

11 

2\  in-. 

n 

i    in. 

10 

2Jin. 

4 

i    in. 

9 

2f  in. 

4 

l.in. 

8 

3    m. 

3i 

IJin. 

7 

l^in. 

7 

Acme  Standard  Thread 

The  Acme  standard  thread  is  an  adaptation  of  the  most 
commonly  used  worm  thread  and  is  intended  to  take  the  place 
of  the  square  thread.  It  is  more  shallow  than  the  worm 
thread  but  was  the  same  depth  as  the  square  thread  and  is 
much  stronger  than  the  latter. 

The  worm  thread  has  sides  with  an  angle  of  29° ;  the  top  is 
flat,  .335  P,  and  the  bottom  is  .31  P.     The  depth  is  .6866  P, 

the  double  depth  being  1.3732  P;  d=D-^"^''^'^  ,  that  is,  the 

N 
diameter  at  the  bottom  of  a  worm  thread   is  equal   to  the 
diameter  of  the  worm  minus  1.3732"  divided  by  the  number 
of  threads  to  one  inch. 


144  VOCATIONAL  MATHEMATICS 

Square  Thread 

A  square   thread  has  parallel   sides;  the  thickness  of  the 
thread  and  its  depth  are  each  one  half  the  pitch :  d  =  D  —  —, 

that  is,  the  diameter  at  the  bottom  of  a  square  thread  is  equal 
to  the  diameter  of  the  screw  minus  one  inch  divided  by  the 
number  of  threads  to  one  inch.  The  thrust  on  the  square- 
threaded  screw  is  parallel  to  the  axis  of  the  screw;  couse- 
quently  the  frictional  losses  are  not  so  great  in  this  form  as 
in  the  V  form,  but  it  is  not  so  strong  in  the  base  as  the  V  thread. 

EXAMPLES 

1.  What  is  the  diameter  of   the  root  of  a  |"  V-threaded 
bolt  with  20  threads  to  the  inch? 

2.  What   is   the   diameter  of  the  root   of   a  |"    U.  S.  S. 
threaded  bolt  with  16  threads  to  the  inch? 

3.  What  is  the  thread  pitch  of  a  V-threaded  screw  with 
20  threads  to  an  inch  ? 

4.  What  is  the  thread  pitch  of  an  Acme  worm  screw  with 
20  threads  to  an  inch  ? 

5.  What  is    the  thread  pitch  of   a  square-threaded  screw 
with  14  threads  to  an  inch  ? 

6.  What  is  the  diameter  of  the  root  of  a  y\"  V-threaded 
bolt  with  18  threads  to  the  inch  ? 

7.  What  is  the  diameter  of   the  root  of  a  |"  V-threaded 
bolt  with  14  threads  to  the  inch  ? 

8.  A  U.  S.  S.  screw  has  20  threads.     What  is  its  thread 
pitch  ?     What  is  its  depth  ? 

9.  What  is  the  diameter  of  the  root  of  a  f "  U.  S.  S.  screw 
with  20  threads  to  the  inch  ? 

10.    What  is  the  thread   pitch  of  a  square-threaded  screw 
with  8  threads  to  the  inch? 


BOLTS,  SCREWS,  AND   RIVETS  145 

11.  A  2 J"  diameter  bolt  has  a  diameter  of  1.962"  at  the  root 
of  the  thread.  What  is  the  depth  of  the  thread  ?  If  the  bolt 
has  4.V  threads  per  inch,  what  is  the  pitch  of  the  threads  ? 

12.  What  is  the  depth  of  a  U.  S.  S.  threaded  screw  of  7 
threads  ? 

13.  What  is  the  diameter  of  the  root  of  a  V-threaded  J" 
screw  with  11  threads  to  the  inch  ? 

14.  What  is  the  depth  of  a  V-threaded  screw  with  11 
threads  to  the  inch? 

15.  What  is  the  diameter  of  the  root  of  a  U.  S.  S.  threaded 
bolt  of  11  threads  ? 

16.  What  is  the  diameter  of  the  root  of  a  U.  S.  S.  J" 
threaded  bolt  of  8  threads? 

17.  What  is  the  thread  pitch  of  a  U.  S.  S.  threaded  screw 
of  9  threads  ? 

18.  What  is  the  depth  of  a  worm-threaded  screw  with  16 
threads  to  the  inch  ? 

19.  What  is  the  depth  of  a  square-threaded  screw  with  8 
threads  to  the  inch  ? 

20.  What  is  the  depth  of  a  U.  S.  S.  threaded  screw  with  8 
threads  to  the  inch  ? 

21.  What  is  the  depth  of  a  U.  S.  S.  1  J"  threaded  bolt  of 
9  threads  ? 

22.  What  is  the  diameter  of  the  root  of  a  U.  S.  S.  threaded 
screw  of  9  threads  ? 

23.  What  is  the  depth  of  a  V-threaded  screw  with  14 
threads  to  the  inch? 

24.  What  is  the  thread  pitch  of  a  V-threaded  screw  with 
16  threads  to  the  inch  ? 

25.  A  U.  S.  S.  threaded  bolt  has  9  threads  to  the  inch. 
What  is  its  thread  pitch  ? 


PART  V  — SHAFTS,   PULLEYS,  AND   GEARING 

CHAPTER   IX 
SHAFTS  AND  PULLEYS 

In  a  machine  shop  one  notices  at  once  the  revolution  of  the 
shafting.  There  is  one  long  cylindrical  bar  called  the  main 
line  attached  to  the  ceiling.  The  power  that  drives  the  ma- 
chinery is  taken  from  this  main  line  by  means  of  pulleys  and 
belts  to  smaller  shafts  called  countershafts.  The  machinery 
is  driven  directly  from  the  countershafts  while  the  main  line 
is  driven  from  a  motor  or  engine  and  flywheel. 

Shafts  of  different  sizes  are  used  according  to  the  horse 
power  required.  ,To  determine  the  horse  power  (H.  P.)  of  a 
shaft,  multiply  the  speed  (revolutions  per  minute)  by  the 
cube  of  the  diameter  of  the  shaft,  and  divide  the  product  by 
84  for  a  steel  shaft  or  by  160  for  an  iron  shaft,  and  the 
quotient  is  the  H.  P.  (For  further  discussion  of  horse  power 
see  pages  188  and  225.) 

EXAMPLES 

1.  What  is  the  H.P.  of  a  2Jg"  steel  shaft  having  280  revo- 
lutions per  minute  ? 

2.  What  is  the  H.P.  of  a  2J"  iron  shaft  having  245  revolu- 
tions per  minute  ? 

3.  What  is  the  H.P.  of  a  2|"  steel  shaft  having  290  revolu- 
tions per  minute  ? 

4.  What  is  the  H.P.  of  a  2f^''  iron  shaft  having  350  revolu- 
tions per  minute  ? 

146 


SHAFTS  AND  PULLEYS  147 


Belting 

Belts  for  transmitting  ix)wer  are  divided  into  two  general 
classes :  leather  belts  and  canvas  belts.  Both  are  sold  by 
the  foot.  The  material  of  the  belt,  the  thickness,  and  width 
determine  its  value.  Coils  of  belting  need  not  be  stretched 
out  to  measure  their  length  (A).  To  do  this  first  count  the 
number  of  coils  (N),  measure  the  diameter  of  the  hole  in  the 
center  of  the  coil  (d),  and  the  outside  diameter  of  the  roll  (/>). 

Then  L  =  0.1309  N(D  +  d) 

In  this  formula  Z=:  length  in  feet,  and  D  and  d  diameter 
in  inches. 

This  rule  is  used  in  estimating. 
The  formula  is  obtained  as  follows : 

— ^^—  =  average  diameter  of  coil 

The  length  L  =  circumference  of  average  diameter 


C  = 

2 


^-12 

—  =  0.1309 
24 

Substituting  this  in  the  formula  ^^(-P  +  <^)  =  0.1309  iV^(Z? -h  d) 

EXAMPLES 

1.  How  many  feet  of  belting  are  there  in  a  coil  that  has  a 
diameter  of  18",  if  the  hole  is  3"  in  diameter  and  there  are  36 
coils? 

2.  How  many  feet  of  belting  are  in  a  coil  that  has  a  16"  di- 
ameter, if  the  hole  is  2^"  in  diameter  and  there  are  38  coils? 


148  VOCATIONAL  MATHEMATICS 

Length  of  Belting  on  Pulleys 

To  find  the  length  of  belting  on  pulleys,  add  together  the 
diameters  of  the  pulleys  in  inches  and  divide  the  sum  by  2. 
Multiply  this  quotient  by  3.25.  Add  this  product  to  twice 
the  distance  in  inches  between  the  centers  of  the  pulleys,  and 
divide  by  12.  The  final  quotient  is  the  length  in  feet  of  the 
belting  on  the  pulleys.  This  is  an  approximation  and  applies 
to  open  belts  only. 

Example.  —  The  diameters  of  two  pulleys  are  24"  and  12" 
respectively,  and  the  distance  between  their  centers  is  108 
inches.     Find  the  length  of  the  belting. 

24"  +  12"  =  36"  18"  X  3.25  =  58.50"  ^J^!!  ^  22.8  feet 

36"  ^  2  =  18"  58.5" +216"-  =  274.5"  22.8'  =  22'  11".  Ans. 

EXAMPLES 

1.  What  is  the  length  of  the  belting  connecting  two  pulleys 
having  diameters  of  18''  and  12"  respectively,  if  the  distance 
between  their  centers  is  92"  ? 

2.  What  is  the  length  of  the  belting  on  two  pulleys  having 
diameters  of  16"  and  22"  respectively,  if  the  distance  between 
their  centers  is  86"  ? 

3.  Find  how  many  feet  of  belting  are  needed  to  make  a  belt 
to  run  over  two  pulleys  each  30"  in  diameter  if  the  distance 
between  their  centers  is  13'. 

Arc  of  Contact 

In  setting  up  machinery  where  there  are  pulleys,  it  is  some- 
times desirable  to  find  the  arc  of  contact  on  the  smaller  pulley. 
The  Boston  Belting  Company  gives  this  rule,  which  holds  when 
the  pulleys  are  nearly  of  the  same  diameter :  Divide  the  dif- 
ference between  the  diameters  of  the  two  pulleys  by  the  dis- 
tance between  the  centers  of  the  shafts,  both  being  in  the  same 


SHAFTS  AND  PULLEYS 


149 


denomination ;  multiply  the  quotient  by  57,  and  subtract  this 
product  from  180 ;  the  result  will  be  the  number  of  degrees 
in  the  arc  of  contact. 

Multiply  the  entire  circumference  of  the  smaller  pulley  in  feet 
by  the  degrees  of  the  arc  of  contact  as  above,  divide  by  360, 
and  the  result  "will  be  the  number  in  feet  of  the  arc  of  contact 
of  the  belt  on  the  smaller  pulley. 


EXAMPLES 

1.  Two  pulleys,  one  18"  and  the  other  24"  in  diameter,  are 
connected  by  a  belt.  If  the  distance  between  the  centers  of 
the  shafts  is  8'  6",  what  is  the  number  of  feet  in  the  arc  of 
contact  of  the  belt  on  the  smaller  pulley  ? 

2.  Find  the  number  of  lineal  inches  that  a  belt  touches  a 
pulley  when  the  arc  of  contact  is  240°,  if  the  diameter  of  the 
pulley  is  4  feet. 

3.  If  the  distance  between  the  centers  of  two  shafts  is  9'  8" 
and  the  pulleys  are  22"  and  16",  what  is  the  number  of  feet  of 
arc  of  contact  of  the  belt  on  the  smaller  pulley  ? 

D 


A  common  way  for  one  shaft  to  drive  another  is  by  means  of 
a  belt  running  upon  two  pulleys,  one  on  the  driving  shaft  and 
the  other  on  the  driven,  as  in  the  above  figure.  In  solving 
problems  the  pulley  on  the  driving  shaft  is  called  the  driver 
and  the  one  on  the  driven  shaft  the  driven. 

In  order  to  install  machinery  and  have  it  run  at  the  proper 
speed,  the  relations  between  the  driving  and  the  driven  pulleys 


150  VOCATIONAL  MATHEMATICS 

and  the  different  methods  of  transmitting  power  from  one 
shaft  to  another  must  be  thoroughly  understood.  To  make 
the  shafts  and  pulleys  run  at  the  proper  speed  the  correct 
diameter  and  circumference  of  the  driving  and  the  driven 
pulleys  must  be  known. 

At  every  revolution  of  the  driver  the  belt  is  pulled  through 
a  distance  equal  to  the  circumference  of  the  driver;  in  moving 
a  distance  equal  to  the  circumference  of  the  driven  pulley,  the 
belt  turns  the  driven  pulley  one  revolution.  When  two  pulleys 
are  connected  by  a  belt,  their  rim  speeds  are  equal.  If  we 
divide  the  distance  the  belt  has  moved  by  the  circumference  of 
the  pulley,  the  quotient  gives  the  number  of  revolutions  of 
the  pulley.  The  smaller  pulley  revolves  at  the  higher  speed, 
a  fact  that  is  usually  stated  mathematically  by  saying  that  the 
revolutions  of  the  pulleys  are  inversely  proportionate  to  their 
circumferences. 

Example.  —  A  pulley  12  inches  in  diameter  makes  300  revo- 
lutions per  minute.  How  fast  is  the  rim  traveling  in  feet  per 
minute  ? 

The  circumference  equals  the  diameter  multiplied  by  3.1416,  or  approx- 
imately 3|.  ■*■ 

12  X  3.1416  =  37.6992  inches  circumference 

l^JiAlil^^  3.1416' feet 
12 

Since  it  is  running  300  revolutions  per  minute, 
300  X  3.1416  =  942.48  feet  per  minute 

It  is  possible  to  express  the  operations  of  the  above  in  a  for- 
mula by  letters : 

Let  D  =  diameter  of  the  pulley  in  inches 

C  =  circumference  of  the  pulley  in  inches 
R  =  revolutions  per  minute  (abbreviated  R.  P.  M.) 
^  =  3.1416 

F=  feet  per  minute  that  the  rim  travels  (circumference 
speed) 


SHAFTS  AND   PULLEYS  151 

Feet  traveled  per  minute  is   equal  to  the  circumference  in 
inches,  multiplied  by  revolutions  per  minute,  and  divided  by  12. 

CR 

That  is,  F—  — — ,  or  substituting  for  C  its  equal  ttZ), 

1  w 

If  we  know  the  value  of  any  two  of  the  quantities  F,  D,  or 
Bf  we  can  find  the  others: 

(I)  P  =  ^  (2)    />  =  ^  (3)   7J  =  1-2,^ 


EXAMPLES 

1.  If  a  ten-inch  pulley  is  making  300  revolutions  per 
minute,  how  fast  is  a  point  on  the  rim  traveling  in  feet  per 
minute  ? 

2.  The  rim  of  a  14"  pulley  is  running  1048  feet  per  minute. 
How  many  revolutions  are  made  per  minute  ? 

3.  A  pulley  18"  in  diameter  makes  375  revolutions  per 
minute.     How  fast  is  the  rim  traveling  in  feet  per  minute  ? 

4.  What  is  the  diameter  of  a  pulley  making  286  revolutions 
per  minute  if  a  point  on  the  rim  is  traveling  984  feet  per 
minute  ? 

5.  A  9"  pulley  is  making  198  revolutions  por  minute.  How 
fast  is  a  point  on  the  rim  traveling? 

6.  A  pulley  32"  in  diameter  is  making  198  revolutions  per 
minute.     How  fast  is  the  rim  traveling  ? 

Speed 

The  speed  of  any  machine  from  the  driving  shaft  or  motor 
may  be  traced  as  follows : 

The  circumference  of  a  pulley  equals  its  diameter  multiplied 
by  3.1416. 


152  VOCATIONAL   MATHEMATICS 

This  may  be  abbreviated  when  C  stands  for  the  circumfer- 
ence and  D  for  the  diameter : 

C  =  3.1416  D  or  7rZ> 

Let      C  =  circumference  of  driver  pulley 
C"  =  circumference  of  driven  pulley 
D  =  diameter  of  driver  pulley 
Z>'  =  diameter  of  driven  pulley 
N  =  revolutions  of  driver  pulley  per  minute 
jV'  =  revolutions  of  driven  pulley  per  minute 

(C  is  read  C  prime.) 

C  X  N  =z  distance  belt  moves  per  minute  on  driver  wheel 
C  X  A^  =  distance  belt  moves  per  minute  on  driven  wheel 
The  distance  represented  by  C  x  N  h  called  rim  speed  of 

driver  wheel. 

The  distance  represented  by  C"  X  N^  is  called  rim  speed  of 

driven  wheel. 

Since  the  surface  speeds  are  equal, 

(1)  CxN^C^xN',        (2)   g  =  ^, 

Since       C  ^-jtB  ovD^N[  ^ ^ D^N' 

C'=^7rD'  UN  N 

If  we  know  any  three  of  the  above  four  quantities,  the 
fourth  can  be  found : 

If  i)  =  ^^'  N==^^ 

N  D 

The  proportion  may  be  easily  remembered  by  noting  that 
the  primes  come  together  as  middle  terms: 

D:D'::N':N 

The  above  formula  may  be  expressed  in  the  form  of  rules: 
Whenever  one  pulley  drives  another  we  have  four  quantities 
to  consider  —  the  diameters  of  the  two  pulleys  and  the  revo- 


SHAFTS  AND   PULLEYS  153 

lutions  per  minute  of  the  two  pulleys.  The  above  equation 
shows  us  that  there  is  a  definite  relation  between  them.  If 
we  know  any  three  of  the  quantities,  we  may  find  the  fourth 
by  transferring  the  factors  of  the  equation. 

NoTK.  —  II  is  often  difficult  to  remember  these  niles.  In  that  case 
draw  a  sketch  and  place  the  given  infonnation  about  each  pulley.  Let  x 
represent  the  unknown  quantity.  Then  multiply  the  two  numbers  known 
about  one  pulley  and  divide  by  the  number  given  in  the  other  pulley. 
The  quotient  will  represent  x. 

Example.  —  Find  the  size  of  a  pulley  on  a  countershaft  that 
runs  120  revohitions  per  minute,  if  the  diameter  of  the  pulley 
on  line  shaft  is  18"  and  runs  180  revolutions  per  minute. 

18  :  Z>'  :  :  120  :  180 
120  D'  =  18  X  180 
2  Z>'  =  54 
D'  =  27 

EXAMPLES 

1.  The  diameter  of  a  pulley  on  the  line  shaft  is  30"  and  it 
runs  158  revolutions  per  minute;  the  countershaft  runs  400 
revolutions  per  minute.  What  is  the  size  of  the  pulley  on  the 
countershaft  ? 

2.  What  is  the  size  of  a  pulley  on  a  countershaft  of  an 
engine  lathe  if  the  diameter  of  the  pulley  on  the  line  shaft  is 
15"  and  it  runs  150  revolutions  per  minute  while  the  counter- 
shaft runs  140  revolutions  per  minute  ? 

3.  What  is  the  size  of  a  pulley  on  a  countershaft  of  a  planer 
if  the  diameter  of  the  pulley  on  line  shaft  is  26"  and  it  runs 
305  revolutions  per  minute  while  the  countershaft  runs  793 
revolutions  per  minute  ? 

4.  What  size  pulley  should  be  placed  on  the  countershaft  of 
a  band  saw  if  the  diameter  of  the  pulley  on  the  main  line  shaft 
is  14",  if  it  runs  250  revolutions  per  minute,  and  the  counter- 
shaft runs  225  revolutions  per  minute  ? 


154  VOCATIONAL  MATHEMATICS 

5.  A  pulley  30"  in  diameter  on  a  main  shaft  running  180 
revolutions  per  minute  is  required  to  drive  a  countershaft  450 
revolutions  per  minute..  What  will  be  the  diameter  of  the 
pulley  on  the  countershaft  ? 

Example.  —  Find  the  number  of  revolutions  tha£  a  counter- 
shaft is  running  if  the  line  shaft  runs  140  revolutions  per  minute, 
and  the  diameter  of  the  pulley  on  it  is  30"  and  the  diameter 
of  the  pulley  on  the  countershaft  12". 

Z>  :  D'  :  :  .V  :  .V 

30  :  12  :  :  N'  :  140 
2N'  =  700 
JV'=360.     Ans. 

EXAMPLES 

1.  Find  the  number  of  revolutions  per  minute  of  a  counter- 
shaft of  an  engine  lathe  if  driven  by  an  18"  pulley  on  the  line 
shaft  which  runs  168  revolutions  per  minute,  if  the  diameter 
of  the  countershaft  pulley  of  lathe  is  12''. 

2.  What  is  the  speed  of  a  countershaft  of  a  speed  lathe,  if 
the  pulley  on  the  main  shaft  is  10"  with  305  revolutions  per 
minute  and  the  diameter  of  the  pulley  on  the  countershaft  is 
4"? 

3.  Find  the  number  of  revolutions  of  a  countershaft  of  a 
band  saw  if  the  pulley  on  the  main  shaft  is  19"  with  215  revo- 
lutions per  minute  and  the  diameter  of  the  pulley  on  the 
countershaft  is  16". 

4.  What  is  the  speed  of  the  countershaft  of  a  cutting-off 
saw  if  the  pulley  on  the  main  shaft  is  15"  with  230  revolutions 
per  minute  and  the  diameter  of  the  pulley  on  the  countershaft 
is  12"? 

5.  A  pulley  30"  in  diameter  making  180  revolutions  per 
minute  drives  a  countershaft  with  a  12"  pulley.  What  is  the 
speed  of  the  countershaft  ? 


SHAFTS  AND  PULLEYS  155 

6.  The  main  driving  pulley  of  an  engine  is  12'  G"  in  diam- 
eter and  makes  96  revolutions  per  minute;  it  is  belted  to  a 
48"  pulley  on  the  main  shaft.     Find  the  speed  of  the  latter. 

Example.  —  The  line  shaft  of  a  machine  shop  runs  120  rev- 
olutions per  minute,  the  diameter  of  the  pulley  on  the  counter- 
shaft is  15",  and  the  countershaft  runs  240  revolutions  per 
minute.     Find  the  size  of  the  pulley  on  the  main  line. 

D:D'::N':.y  |=^ 

D  :  1 V  :  :  240  :  120 
120 Z>  =  lo''  x240 

D  =  30".     Ans.  N'D'  =  ND 

EXAMPLES 

1.  The  main  line  runs  160  revolutions  per  minute;  the 
countershaft  pulley  is  9"  in  diameter  and  runs  320  revolutions 
per  minute.  What  is  the  diameter  of  the  pulley  on  the  main 
line  ? 

2.  A  pulley  24"  in  diameter  running  144  revolutions  per 
minute  is  to  drive  a  shaft  192  revolutions  per  minute.  What 
must  the  diameter  of  the  pulley  be  on  the  driven  shaft  ? 

3.  A  driving  shaft  runs  140  revolutions  per  minute;  the 
driven  pulley  is  10"  in  diameter  and  is  to  run  350  revolutions 
per  minute.    What  must  the  diameter  of  the  driving  pulley  be  ? 

4.  A  countershaft  with  a  12"  pulley  runs  450  revolutions 
per  minute ;  the  revolutions  of  the  main  shaft  are  180.  What 
size  pulley  must  be  used  on  the  main  shaft  ? 

5.  A  main  line  runs  189  revolutions  per  minute  and  the 
countershaft  pulley  is  10"  in  diameter  and  runs  385  revolutions 
per  minute.     What  is  the  size  of  the  pulley  on  the  main  line  ? 

6.  A  pulley  36"  in  diameter  running  168  revolutions  per 
minute  is  to  drive  a  shaft  212  revolutions  per  minute.  What 
must  be  the  diameter  of  the  pulley  on  the  driven  shaft  ? 


156  VOCATIONAL  MATHEMATICS 

7.  A  driving  shaft  runs  184  revolutions  per  minute;  the 
driven  pulley  is  12"  in  diameter  and  is  to  run  350  revolutions 
per  minute.    What  must  be  the  diameter  of  the  driving  pulley  ? 

8.  What  is  the  speed  of  the  main  shaft  if  the  pulley  is  12" 
and  the  revolutions  per  minute  on  the  other  shaft  are  228  with 
a  pulley  8"  ? 

9.  What  is  the  speed  of  a  countershaft  if  the  pulley  is  11" 
and  the  main  shaft  runs  196  revolutions  per  minute  with  a 
14"  pulley  ?  • 

10.  A  14'  flywheel  running  98  revolutions  per  minute  drives 
a  9"  pulley.     What  is  the  speed  of  the  pulley  ? 

Countershafts  or  Jackshafts 

The  first  shaft  belted  off  from  a  flywheel  is  often  called  a 
jackshaft.  In  the  figure  below  the  jackshaft  carries  the 
pulley  d ;  on  the  main  line  is  the  large  pulley  D  and  the  small 
pulley  F  on  the  jackshaft.  On  another  main  line  is  the  pulley 
/.  Pulley  B  is  the  first  driver  and  by  means  of  pulley  F  on 
the  jackshaft  drives  pulley  d.  Pulley  d  is  the  second  driver. 
Pulley  F  is  the  first  driven  and  pulley /is  the  second  driven 
pulley. 


D 


Main  Line  Jackshaft 

Example.  —  The  main  shaft  runs  160  R.  P.  M.  (revolations 
per  minute) ;  D  is  60"  in  diameter  and  drives  F  140  E,.  P.  M. 
What  is  the  diameter  of  2^? 

The  second  main  /  is  to  run  186  R.  P.  M.  What  should  be 
the  diameter  of/  if  the  72"  driver  d  is  running  140  R.  P,  M,  ? 


SHAFTS  AND  PULLEYS 


157 


An  examination  of  the  data  in  connection  with  D  and  /  in 
the  figure  above  will  show  that  the  relative  speed,of /to  D  is 

XxDxd=^N'xFxf 
where  N=  number  of  revolutions  of  driver 

and  N'  =  number  of  revolutions  of  driven 

That  is,  the  continued  product  of  the  speed  of  the  first  driver 
and  the  diameters  of  all  the  drivers  is  equal  to  the  continued 
product  of  the  speed  of  the  last  driven  by  the  diameters  of  all 
the  driven  pulleys.  In  this  combination  of  driving  f  hy  D 
there  are  six  quantities,  any  of  which  can  be  found  when  we 
know  the  other  five,  by  figuring  from  one  shaft  to  the  next 
step  by  step. 

EXAMPLES 

1.  The  R.P.  M.  of  a  26"  driving  pulley  is  270.     What  are 
the  revolutions  per  minute  of  an  18"  driven  pulley  ? 

2.  If  there  is  a  shaft  with 
a  speed  of  270  R.  P.  M.  upon 
which  there  is  a  26"  pulley 
driving  a  16"  pulley,  and  on  the 
shaft  with  the  16"  pulley  is  a 
30"  pulley  driving  an  18"  pulley, 
what  is  the  speed  of  the  shaft 
which  carries  the  18"  pulley  ? 

3.  A  flywheel  which  is  30  feet  in  diameter  drives  a  coun- 
tershaft by  means  of  a  pulley 
6  feet  in  diameter ;  the  flywheel 
makes  50  R.  P.  M.  What  size 
of  pulley  must  be  used  on  the 
countershaft  to  give  300  R.  P.  M. 
to  a  pulley  2  feet  in  diameter? 

4.  If  3  flywheels,  respectively  13',  lOf ,  and  9'  di- 
ameter have  the  same  circumferential  speed  of  2750' 
per  minute,  how  many  revolutions  per  minute  does  each  make  ? 


270 


158  VOCATIONAL  MATHEMATICS 

5.  A  flywheel  which  is  40  feet  in  diameter  drives  a  counter- 
shaft by  means  of  a  pulley  6^  feet  in  diameter  ;  the  flywheel 
makes  54  R.  P.  M.  What  size  of  pulley  must  be  used  on  the 
countershaft  to  give  341  R.  P.  M.  to  a  pulley  3  feet  in 
diameter? 

6.  The  size  of  a  main  driving  pulley  is  20  feet  in  diameter 
on  a  shaft  with  a  speed  of  70  K.  P.  M.,  driving  a  pulley  4  feet 
in  diameter  and  two  other  pulleys  5.6  and  G.36  feet  in  diameter, 
respectively.     What  is  the  speed  of  each  shaft  ? 


CHAPTER   X 


GEARING 


In  a  machine  shop  power  is  transmitted  from  one  part  of  a 
machine  to  another  part  by  means  of  tooth-shaped  interlocking 
wheels.  The  train  of  toothed  wheels  for  transmitting  motion 
is  called  gearing. 

Gears  are  nothing  but  pulleys  with  teeth,  and  are  made  to 
drive  one  another  by  the  teeth  coming  in  contact  with  each 
other.  If  a  small  gear  drives  a 
larger  gear,  the  larger  gear  will  go 
more  slowly  than  the  smaller ; 
that  is,  the  larger  gear  will  make 
fewer  turns  in  a  minute.  Just 
the  reverse  is  true  if  a  larger  gear 
drives  a  smaller  one.  The  num- 
ber of  revolu- 
tions which 
a  gear  makes 
is  always  pro- 
portional to 
the  number 
of  its  teeth. 

As  in  pul- 
leys,     there 

are  the  driver  and  the  driven  gears.  The  driver  may  be  dis- 
tinguished from  the  driven  by  examining  the  gears  and  notic- 
ing that  it  is  the  gear  that  is  bright  or  worn  on  the  front  of  the 
tooth  —  that  is,  the  side  of  the  wheel  moving.  The  driven  wheel 
is  worn  on  the  side  away  from  the  direction  of  the  motion. 


Bkvel  Gear 


Spur  6bab 


160  VOCATIONAL  MATHEMATICS 

In  order  to  express  the  relation  between  the  driver  and 
driven  gears  it  is  necessary  to  use  abbreviations. 

Let  Z)  =  number  of  teeth  in  driver 
D'  =  number  of  teeth  in  driven 
N=  number  of  revohitions  of  the  driver 
N"  =  number  of  revolutions  of  the  driven 
D:D'  ::  N':  ^Vor  DN=  D'N' 

That  is,  the  product  of  the  teeth  in  the  driver  by  its  revolutions 
equals  the  tooth  transits  of  the  driver,  which  in  turn  equal  the 
tooth  transits  of  the  driven  or  follower.  If  any  three  of  these 
quantities  are  known,  the  fourth  can  be  found. 

Example.  —  How  many  revolutions  does  a  12-tooth  follower 
make  to  five  revolutions  of  a  24-tooth  driver  ? 

12iV^=24x6 
iV=10    Ans. 

EXAMPLES 

1.  A  driver  has  98  teeth  and  its  follower  42.  How  many 
revolutions  will  the  follower  make  to  one  revolution  of  the 
driver  ? 

2.  In  Example  1  how  many  revolutions  of  the  driver  will 
drive  the  follower  one  revolution  ? 

3.  How  many  teeth  must  a  follower  have  in  order  to  make 
three  revolutions  while  a  96-tooth  driver  makes  one  ? 

4.  How  many  teeth  must  a  gear  have  to  revolve  16  times 
while  a  60-tooth  mate  revolves  12  times  ? 

5.  A  96-tooth  gear  drives  a  48-tooth  gear.  What  is  the  ratio 
of  their  speeds  ? 

6.  A  48-tooth  gear  drives  a  120-tooth  gear.  What  is  the  ratio 
of  their  speeds  ? 

7.  Two  shafts  are  connected  by  gears,  one  of  which  turns  55 
times  a  minute  and  the  other  11  times  a  minute.  If  the  small 
gear  has  32  teeth,  how  many  teeth  has  the  larger  gear  ? 


GEARING 


161 


Pitch 

In  order  to  solve  problems  connected  with  the  use  of  gears 
it  is  necessary  to  know  the  different  terms  used  in  connection 
with  gearing. 

The  pitch  circle  is  shown  in  the  illustration  and  is  the  circle 
which  runs  around  the  teeth.     It  is  the  same  size  as  the  fric- 


Gear  Teethe 


L  Pitch 

r^    lAat 


Diameter 


tion  rollers  or  cylinders  would  be  if  no  teeth  were  there. 
When  two  spur  gears  roll  together  their  pitch  circles  are  con- 
sidered to  be  always  in  contact.  The  jy^tch  diameter  is  the 
diameter  of  the  pitch  circle.  The  word  diameter  when  applied 
to  gears  always  means  the  pitch  diameter. 

The  circular  pitch  is  the  distance  measured  on  the  pitch 
circle  from  the  center  line  of  one  tooth  to  the  center  of  the 
next.  This  is  illustrated  in  the  diagram.  In  solving  gearing 
problems  the  circular  pitch  is  not  nearly  so  important  as  the 
diametral  pitch. 

If  the  distance  from  the  center  of  a  tooth  to  the  center  of  tlie  next 
tooth  is  i",  the  gear  is  I"  circular  pitch. 

The  diametral  pitch  is  the  number  of  teeth  for  each  inch  of 
pitch  diameter. 

For  example,  if  a  gear  has  thirty  teeth  and  the  pitch  diameter  is  three 
inches,  then  the  diametral  pitch  is  .SO  -=-  .3  =  10,  or  a  10  diametral  pitch 
gear.  Using  P  for  the  diametral  pitch,  D  for  the  pitch  diameter,  and  ^V 
for  the  number  of  teeth,  we  liave  a  formula  P  =  N  ^  D. 


162  VOCATIONAL  MATHEMATICS 

To  find  the  thickness  of  tooth  at  the  pitch  line  when  the 
diametral  pitch  is  given,  divide  1.57  by  the  diametral  pitch. 

For  example,  if  the  diametral  pitch  is  3,  divide  1.57  by  3,  and  the  quo- 
tient, which  is  .523  inches,  is  the  thickness  of  the  tooth. 

To  find  the  circular  pitch  when  the  diametral  pitch  is  given 
divide  3,1416  by  the  diametral  pitch. 

For  example,  if  the  diametral  pitch  is  4,  divide  3.1416  by  4  and  the 
quotient,  which  is  .7854,  is  the  circular  pitch. 

Diametral  pitch  is  found  when  the  circular  pitch  is  given  by 
dividing  3.1416  by  the  circular  pitch.  Since  the  circumference 
of  any  circle  is  equal  to  3.1416  times  its  diameter,  every  inch  of 
diameter  of  any  circle  is  equal  to  3.1416  inches  of  circum- 
ference, or  in  this  instance,  for  every  inch  of  pitch  diameter 
we  have  3.1416  inches  of  circumference  measured  on  the 
pitch  circle.  But  the  diametral  pitch  is  by  definition  the  num- 
ber of  teeth  for  each  inch  of  pitch  diameter,  and  by.  the  above 
reasoning  we  can  also  say  that  the  diametral  pitch  is  the  num- 
ber of  teeth  for  each  3.1416  inches  of  circumference  of  pitch 
circle. 

Since  the  circular  pitch  is  the  distance  from  the  center  of 
one  tooth  to  the  center  of  the  next,  it  will  also  be  equal  to 
3.1416  of  circumference  of  pitch  circle  divided  by  the  number 
of  teeth  in  that  3.1416.  But  tlie  number  of  teeth  in  3.1416  is 
equal  to  the  diametral  pitch,  and  therefore  the  distance  from 
the  center  of  one  tooth  to  the  center  of  the  next,  or  the  circular 
pitch,  is  equal  to  3.1416  divided  by  the  diametral  pitch. 
Therefore,  using  P  for   circular   pitch  and  P  for   diametral 

pitch,  P  =  3.1416  -  F  or  P'  =  5:111^. 

EXAMPLES 

1.  What  is  the  diametral  pitch  of  a  gear  liaviug  (a)  56  teeth 
and  a  pitch  diameter  of  8"  ?  [b)  60  teeth  and  a  pitch  diameter 
of  12"  ? 


GEARING 


163 


2.  What  is  the  thickness  of  the  tooth  of  a  gear  having 
diametral  pitch  of  (a)  8?  (6)  4?     (c)  14? 

3.  What  is  the  eirciiUir  pitch  if  the  diametral  pitch  of 
geAris(a)  8?     (6)  10?     (c)  5? 

4.  What  is  the  diametral  pitch  if  the  circular  pitch  of 
gear  is  (a)  1.5708?     (b)  .5230?     (c)  .2618*^     ((J)  .7854? 

To  find  the  pitch  di- 
ameter when  the  number 
of  teeth  and  the  diame- 
tral pitch  aregiven, divide 
the  number  of  teeth  by 
the  diametral  pitch.  For 
example,  if  the  number 
of  teeth  is  40  and  the 
diametral  pitch  is  4, 
divide  40  by  4,  and 
the  quotient,  which  is 
10",  is  the  pitch  di- 
ameter. 

To  make  a  gear  the 
outside  diameter  must 
first  be  known.  If  the 
diametral  pitch  and  the 
number  of  teeth  are 
given,  the  outside  di- 
ameter may  be  found 
by  the  following  rule: 


D  = 


N-\-2 


where  D  =  outside  di- 
ameter, N=  number  of 
teeth,  and  P=  diametral 
pitch. 


164  VOCATIONAL  MATHEMATICS 

For  example,  if  a  gear  has  24  teeth  and  its  diametral  pitch  is  2,  the 

outside  diameter  will  be  13";  O  =  ^£+2  ^  26  ^  ^g,,    ^^^      ^  :^±2. 

2  2'  0 

To  find  the  diame^raZj[)iYc7i  when  the  number  of  teeth  and  the 
diameter  of  the  blank  are  given,  add  2  to  the  number  of  teeth 
and  divide  by  the  diameter  of  the  blank. 

For  example,  if  the  number  of  teeth  is  40  and  the  diameter  of  the  blank 
is  10^",  add  2  to  the  number  of  teeth,  making  42,  and  divide  by  10^,  and 
the  quotient,  which  is  4,  is  the  diametral  pitch. 

To  find  the  thickness  of  tooth  at  the  pitch  line  when  the  cir- 
cular pitch  is  given,  divide  the  circular  pitch  by  2. 

For  example,  if  the  circular  pitch  is  1.047",  divide  this  by  2,  and  the 
quotient,  which  is  .523,  is  the  thickness  of  tooth. 

To  find  the  number. of  teeth  when  the  pitch  diameter  and  the 
diametral  pitch  ai-e  given,  multiply  the  pitch  diameter  by  the 
diametral  pitch. 

For  example,  if  the  pitch  diameter  is  10"  and  the  diametral  pitch  is  4, 
multiply  10  by  4,  and  the  product  40  will  be  the  number  of  teeth  in  the 
gear. 

To  find  the  whole  depth  of  tooth  divide  2.157  by  the  diametral 
pitch. 

For  example,  if  the  diametral  pitch  is  6,  divide  2.157  by  6,  and  the 
quotient,  .3595",  is  the  whole  depth  of  the  tooth. 

Formulas  for  Determining  the  Dimensions  of  Gears  bt 
Diametral  Pitch 

P  =  the  diametral  pitchy  or  the  number  of  teeth  to  one  inch  of 

diameter  of  pitch  circle. 
D'  =  the  diameter  of  pitch  circle. 

D  =  the  whole  (outside)  diameter  or  the  diameter  of  the  blank. 
JV  =  the  number  of  teeth. 
V=  the  velocity. 

t  =  the  thickness  of  tooth  or  cutter  on  pitch  circle. 
D"  =  the  working  depth  of  tooth. 


GEAKING  165 

/  =  the  amount  added  to  depth  of  tooth  for  rounding  the  corners 
and  for  clearance. 
D"  4-/  =  the  whole  depth  of  tooth. 
r  denotes  the  constant  3.1416. 

P'  denotes  the  circular  pitch  or  the  distance  from  the  center  of  one  tooth 
to  the  center  of  the  next  on  the  pitch  circle. 

FormuUm  KjrnmpleH ' 

^±1  =         71+^,  or  T?+_2_  12. 

6.166  '         6fV 

=        1:^^  =  12. 
6 

6.166  X  72  ^g 
72  +  2 

=        ^  =  6. 
12 

=        12  X  6  =  72. 

=         12  X  6.166  -  2,  or  12  X  6j-\  -  2  =  72. 

72  +  2 


D 

P-- 

N 

=  — 

D' 

DxX 

N-^2 

D' 

N 
P 

X. 

=  PD^ 

X. 

=  PD-2 

n. 

_.V+2 

P  12 


rr 


=  6  166,  or  (jj\. 


D=D'-h-  =         6  +  :p^,or  6  +  .166=6.166. 

^  =  1:51  =         M[  =  .130.      • 
P  12 

9  O  O 

f  =  —  =        :^  =  .013. 
10  10 

/>"+/  =        .166 +  .013  =  .179. 

rw         TT  3.1416  nf,n 

^  =  P  =        -12- =  •2^2- 


3.1416 


P  =  ^  =        ^•"":^  =  12. 

F  .262 

1  The  examples  placed  opposite  the  formulas  above  are  for  a  single  gear  of 
12  pitch,  6.166"  or  Gft"  diameter. 


166  VOCATIONAL  MATHEMATICS 

EXAMPLES 

Solve  by  using  the  above  fonnulas  and  the  rules  on  pages 
161-165 : 

1.  If  the  number  of  teeth  of  a  gear  is  46  and  the  diametral 
pitch  is  8,  what  will  the  pitch  diameter  be  ? 

2.  If  a  gear  is  to  have  54  teeth  and  the  diametral  pitch  is  6, 
what  will  be  the  pitch  diameter  ? 

3.  If  a  gear  has  38  teeth  and  the  diametral  pitch  is  6,  what 
is  the  pitch  diameter  ? 

4.  If  the  pitch  diameter  of  a  gear  is  6"  and  the  diametral 
pitch  is  8,  what  will  be  the  number  of  teeth? 

5.  How  many  teeth  has  a  gear  if  the  pitch  diameter  is  7f " 
and  the  diametral  pitch  is  6  ? 

6.  How  many  teeth  has  a  gear  if  the  pitch  diameter  is  8J" 
and  the  diametral  pitch  is  4? 

7.  If  the  pitch  diameter  of  a  gear  is  12"  and  the  diametral 
pitch  7,  how  many  teeth  has  the  gear  ? 

8.  If  the  diametral  pitch  is  16  and  the  pitch  diameter  of  a 
gear  is  6J",  how  many  teeth  has  the  gear  ? 

9.  What  is  the  whole  depth  of  the  tooth  if  the  diametral 
pitch  of  a  gear  is  (a)  8?  .  (6)  4  ?     (c)  7  ?     (d)  10? 

10.  If  a  gear  has  48  teeth  and  the  diameter  of  the  blank 
is  6y,  find  the  diametral  pitch. 

11.  If  a  gear  has  61  teeth  and  the  diameter  of  the  blank 
is  101",  find  the  diametral  pitch. 

12.  If  a  gear  has  35  teeth  and  the  diameter  of  the  blank 
is  Syy,  find  the  diametral  pitch. 

13.  If  a  gear  has  78  teeth  and  the  diameter  of  the  blank 
is  5",  find  the  diametral  pitch. 

14.  If  a  gear  has  54  teeth  and  the  diameter  of  the  blank 
is  6|",  find  the  diametral  pitch. 

15.  What  is  the  thickness  of  the  tooth  if  the  circular  pitch 
of  a  gear  is  (a)  .1848"  ?     (b)  .1428"  ?     (c)  6.2832"  ?     (d)  2"  ? 


GEARING  167 

If  one  examines  closely  the  movements  of  two  gears  on  a 
machine,  one  will  notice  that  all  the  teeth  pass  a  given  point  at 
every  revolution  of  the  wheel.     That 
is,  the  teeth  marked   Tty  on  the  gear  f^J  lyO 

shown  in  the  illustration  pass  an  out-  ^  ^) 

side  point  like  P  in  making  one  com-    2'— CT  7^^ 

plete  revolution.     So  when  a  tooth  and  (\  /O— * 

the  space  adjacent  to  it  have  passed  a  (j^  r\j 

given  point  like  P,   one   transit   of  a 

tooth  has  occurred.     There  are  as  many  tooth  transits  at  every 
complete  revolution  of  a  gear  as  there  are  teeth  in  the  gear. 

If  we  multiply  the  number  of  teeth  by  the  number  of 
revolutions,  the  product  will  be  the  number  of  transits.  If 
this  product  (number  of  transits)  be  divided  by  the  number 
of  teeth,  the  quotient  will  be  the  revolutions  of  the  gear.  For 
example,  if  a  12-tooth  gear  makes  CO  transits,  then  it  has  made 
(^J  =  5)  five  revolutions. 


Two  shafts,  D  and  Fy  are  to  be  connected  by  gears  so  that 
shaft  D  will  make  one  revolution  while  shaft  F  makes  two. 
To  do  this  a  gear  must  be  put  on  shaft  D  having  twice  the 
number  of  teeth  that  the  gear  on  shaft  F  has.  If  a  gear  hav- 
ing 24  teeth  is  put  on  shaft  D,  the  gear  on  shaft  F  will  have 
half  as  many.     Each  time  the  gear  on  D  turns  around  once  the 


168  VOCATIONAL  MATHEMATICS 

gear  on  F  will  turn  twice ;  that  is,  the  24  teeth  on  gear  D  will 
have  to  turn  gear  F  twice  in  order  to  mesh  with  the  24  teeth 
on  F. 

The  relation  or  ratio  of  the  speed  of  F  to  the  speed  of  D  is 
2  to  1.  This  is  called  the  ratio  of  the  gearing.  The  ratios 
between  the  speeds  and  number  of  teeth  can  be  written  in  the 
form  of  a  proportion  : 

24:12::  2:1 

A^nd  the  number  of  teeth  on  gear  D  is  to  the  number  of  teeth 
on  gear  F  as  the  speed  of  F  is  to  the  speed  of  Z>. 


Train  of  Gears 

When  two  gears  mesh,  as  in  the  illustration  on  page  167,  one 
revolves  in  the  opposite  direction  from  the  other.  Three  or 
more  gears  running  together,  as  in  the  illustration  above,  are 
often  called  a  train  of  gears.  In  a  train  of  spur  gears  like 
these,  one  gear,  J,  which  is  called  an  intermediate  gear,  meshes 
with  the  other  two  gears,  D  and  F,  and  causes  the  revolutions 
of  both  Z)  and  F  to  be  made  in  one  direction,  while  the  inter- 
mediate revolves  in  the  opposite  direction.  The  intermediate 
does  not  change  the  relative  speeds  of  D  and  F,  so  that  they 
can  be  figured  as  explained  on  page  167.  An  intermediate 
gear  is  also  called  an  idler. 


GEARING  169 

We  may  calculate  the  nuinbsr  of  revolutions  of  any  follower 
for  any  number  of  revolutions  of  a  driver,  in  a  train  of  gears, 
step  by  step.  To  find  the  number  of  revolutions  of  the  last 
follower  when  the  number  of  revolutions  of  the  first  driver 
and  the  teeth  in  all  the  gears  are  known,  take  the  continued 
product  of  the  revolutions  of  the  first  driver  and  all  the  driv- 
ing gears,  and  divide  it  by  the  continued  product  of  all  the 
followers ;  the  quotient  is  the  number  of  revolutions  of  the  last 
follower.  In  other  words,  the  product  of  the  revolutions  of 
the  first  driver  and  the  teeth  of  all  the  driving  gears  is  equal 
to  the  continued  product  of  the  revolutions  of  the  last  follower 
and  the  teeth  of  all  the  driven  gears. 

EXAMPLES 

1.  If  a  60-tooth  wheel  is  to  mesh  with  one  having  46  teeth 
and  the  60-tooth  gear  makes  25  revolutions  per  minute,  how 
many  will  the  46-tooth  gear  make  ? 

2.  A  168-tooth  gear  drives  a  28-tooth  gear.  What  is  the 
ratio  of  the  gearing  ? 

3.  What  would  be  the  ratio  of  the  gearing  in  the  example 
above  if  the  28-tooth  gear  were  the  driver  ? 

4.  If  the  28-tooth  gear  is  making  48  revolutions  per  min- 
ute, how  many  revolutions  per  minute  is  the  168-tooth  gear 
making  ? 

5.  How  could  the  gearing  be  changed  to  make  the  28-tooth 
gear  turn  in  the  same  direction  as  the  168-tooth  gear  ? 

6.  Two  gears  running  together  have  a  speed  ratio  of  7  to  1. 
If  the  smaller  one  turns  14  times,  how  many  times  will  the 
larger  one  turn  ? 

7.  If  a  144-tooth  gear  makes  one  complete  turn,  how  many 
turns  will  a  32-tooth  gear  make  working  with  it  ? 

8.  In  the  above  example  if  the  32-tooth  gear  turned  once, 
how  many  turns  will  the  144-tooth  gear  make  ? 


170  VOCATIONAL  MATHEMATICS 

9.  A  train  of  3  gears  has  69,  30,  and  74  teeth.  If  the  69- 
tooth  gear  makes  100  revolutions  per  minute,  how  many 
revolutions  per  minute  will  the  74-tooth  gear  make  ?  Make 
a  sketch  showing  the  direction  in  which  each  gear  turns. 

10.  A  train  of  gears  is  made  up  of  6  gears  having  teeth  as 
follows:  46,  60,  32,  72,  56,  and  48;  while  the  first  gear  in  the 
train  makes  10  turns,  how  many  turns  will  the  last  gear  make  ? 

11.  What  two  gears  will  give  a  ratio  of  speeds  so  that  the 
driver  will  make  if  as  many  turns  as  the  follower ;  that  is, 
while  the  driver  makes  13  turns  the  follower  will  make  14  ? 

12.  If  24  gears  work  in  a  train,  in  what  direction  will  the 
last  one  turn  if  the  first  turns  right-handed  ? 

REVIEW   EXAMPLES 

1.  Two  gears  working  together  have  pitch  circles  14^'  and 
26f "  in  diameter.    What  is  the  distance  between  their  centers  ? 

The  distance  between  centers  may  be  obtained  from  the  following 

formulas : 

D'  -\-d'  h 

«=  — ^r — '•<'''  =  7r^ 
2  2P 

where  a  —  distance  between  centers 

d'  =  diam.  of  pitch  circle  of  smaller  gear 

h  =  number  of  teeth  on  both  wheels 

2.  Two  gears  in  mesh  have  pitch  circles  17.082"  and 
31.3124"  in  diameter.     What  is  their  center  distance? 

3.  Two  gears  working  together  are  four  pitch,  the  larger 
has  36  teeth  and  the  smaller  has  14  teeth.  What  is  their 
center  distance  ? 

4.  The  circular  pitch  of  a  planer  table  gear  is  1".  What 
should  be  the  total  depth  to  cut  the  tooth  space  ? 

5.  The  circular  pitch  of  a  milling  machine  gear  is  |^"  and  it 
has  96  teeth.     What  is  its  outside  diameter  ? 

6.  A  flue  rattler  gear  is  to  be  renewed.  Its  outside  diam- 
eter is  34"  and  it  has  100  teeth.  What  size  shall  we  draw  its 
pitch  circle  ? 


GEARING  171 

7.  Two  gears  mesh  together,  one  has  28  teeth  and  an 
outside  diameter  of  2|"  and  the  other  68  teeth  and  an  outside 
diameter  of  5|".     Find  the  center  distance. 

8.  A  gear  has  a  circular  pitcli  of  |f ".  What  is  the  thick- 
ness of  the  tooth  at  the  pitch  line  ? 

9.  A  pitch  circle  is  18"  in  circumference.  If  the  teeth  are 
I"  thick,  what  is  the  circular  pitch  ? 

10.  A  gear  is  10  pitch.  What  is  the  total  depth  of  its 
teeth  ? 

11.  The  diametral  pitch  of  a  gear  is  4.  What  is  the  cir- 
cular pitch  ? 

12.  The  circular  pitch  of  a  gear  is  .157.  What  is  the  di- 
ametral pitch  ? 

13.  A  gear  is  12  pitch.     What  is  its  circular  pitch  ? 

14.  A  gear  is  20  pitch.  If  it  has  105  teeth,  what  is  its  out- 
side diameter  ? 

15.  A  gear  of  10  pitch  has  44  teeth.  Find  the  pitch 
diameter. 

16.  A  gear  has  28  teeth  and  a  pitch  diameter  of  8.  What 
is  the  pitch  ? 

17.  A  gear  is  set  in  a  milling  machine  ready  to  cut  the 
teeth,  and  if  the  pitch  is  to  be  one,  what  is  the  depth  of  the 
cut? 

la  A  gear  is  14  pitch  and  has  an  outside  diameter  of  4|". 
Find  the  depth  of  cut  for  the  teeth. 

19.  A  gear  has  75  teeth  and  is  18  pitch.  What  is  the  work- 
ing depth  of  the  tooth  ? 

20.  A  pinion  has  44  teeth  and  is  10  pitch.  What  is  the  out- 
side diameter  ? 

21.  A  lathe  back  gear  has  108  teeth  and  is  5  pitch.  What 
is  its  outside  diameter  ? 


PART   VI  — PLUMBING   AND   HYDRAULICS 


CHAPTER   XI 

RECTANGULAR  AND  CYLINDRICAL  TANKS 

Water  Supply.  —  The  question  of  the  water  supply  of  a  city  or  a  town 
is  very  important.  Water  is  usually  obtained  from  lakes  and  rivers 
which  drain  the  surrounding  country.  If  a  lake  is  located  in  a  section 
of  the  surrounding  country  higher  than  the  city  (which  is  often  located 
in  a  valley),  the  water  may  be  obtained  from  the  lake,  and  the  pressure 
of  the  water  in  the  lake  may  be  sufficient  to  force  it  through  the  pipes 


'm&WW' 


Water  Supply 


into  the  houses.  But  in  most  cases  a  reservoir  is  built  at  an  elevation  as 
high  as  the  highest  portion  of  the  town  or  city,  and  the  water  is  pumped 
into  it.  Since  the  reservoir  is  as  high  as  the  highest  point  of  the  town, 
the  water  will  flow  from  it  to  any  part  of  the  town.  If  houses  are 
built  on  the  same  hill  with  the  reservoir,  a  stand-pipe,  which  is  a  steel 
tank,  is  erected  on  this  hill  and  the  water  is  pumped  into  it. 

Water  is  conveyed  from  the  reservoir  to  the  house  by  means  of  iron 
pipes  of  various  sizes.  It  is  distributed  to  the  different  parts  of  the  house 
by  small  lead,  iron,  or  brass  pipes.  Since  water  exerts  considerable  pres- 
sure, it  is  necessary  to  know  how  to  calculate  the  exact  pressure  in  order 
to  have  pipes  of  proper  size  and  strength. 

172 


PLUMBING  AND  HYDRAULICS  173 

EXAMPLES 

1.  Water  is  measured  by  means  of  a  meter.  If  a  water 
meter  measures  for  tive  hours  7()0  cubic  feet,  how  many  gal- 
lons does  it  indicate  ? 

Note.  — 231  cubic  inches  =  1  gallon. 

2.  If  a  water  meter  registered  1845  cubic  feet  for  3  days, 
how  many  gallon?  were  used  ? 

3.  A  tank  holds  exactly  12,852  gallons  ;  what  is  the  capacity 
of  the  tank  in  cubic  feet  ? 

4.  A  tank  holds  3841  gallons  and  measures  4  feet  square  on 
the  bottom  ;  how  high  is  the  tank  ? 

Rectangular  Tanks.  —  To  find  the  contents  in  gallons  of  a  square  or 
rectangular  tank,  multiply  together  the  length,  breadth,  and  height  in 
feet ;  multiply  the  result  by  7.48. 

I  =  length  of  tank  in  feet 
b  =  breadth  of  tank  in  feet 
h  =  height  of  tank  in  feet 
Contents  =  Ibh  cubic  feet  x  7.48  =  7.48  Ibh 
gallons 

(Note.  —  1  cu.  ft.  =  7.48  gallons.) 

If  the  dimensions  of  the  tank  are  in  inches,  multiply  the  length, 
breadth,  and  height  together,  and  the  re.sult  by  .004829. 

5.  Find  the  contents  in  gallons  of  a  rectangular  tank  having  in- 
side dimensions  (a)  12'  x  8'  x  8';  (b)  15"  x  11"  X  6" ;  (c)  3'  4" 
X2'8"x8";  (d)  5'8"x4'3"x3'5";  (e)  3' 8"  x  3' 9"  x  2' 5". 

Cylindrical  Tank.  —  To  find  the  contents  of  a  cylin- 
drical tank,  square  the  diameter  in  inches,  multiply 
this  by  the  height  in  inches,  and  the  result  by  .0034. 

d  =  diameter  of  cylinder 
h  =  height  of  cylinder 
Contents  =  d^h  cubic  inches  x  .0034  =  d^h  .0034  gallons 


6.    Find  the  capacity  in  gallons  of  a  cylindrical  tank  (a)  14" 
in  diameter  and   8'  high;    (b)  (>"  in  diameter  and   5'   high; 


174  VOCATIONAL  MATHEMATICS 

(c)  15"  in  diameter  and  4'  high;    (d)  1'  8"  in  diameter  and 
5'  4"  high;  (e)  2'  2"  in  diameter  and  6'  V  high. 

Inside  Area  of  Tanks. — To  find  the  area,  for  lining  purposes,  of  a 
square  or  rectangular  tank,  add  together  the  widths  of  the  four  sides  of 
the  tank,  and  multiply  the  result  by  the  height.  Then  add  to  the  above 
the  area  of  the  bottom.  Since  the  top  is  usually  open,  we  do  not  line 
it.     In  the  following  problems  find  the  area  of  the  sides  and  bottom. 

7.  Find  the  amount  of  zinc  necessary  to  'line  a  tank  whose 
inside  dimensions  are  2'  4"x  10"  x  10". 

8.  Find  the  amount  of  copper  necessary  to  line  a  tank 
whose  inside  dimensions  are  1'9"  x  11"  x  10",  no  allowance 
made  for  overlapping. 

9.  Find  the  amount  of  copper  necessary  to  line  a  tank 
whose  inside  dimensions  are  3'  4"  x  1'  2"  x  11",  no  allowance 
for  overlapping. 

10.  Find  the  amount  of  zinc  necessary  to  line  a  tank 
2'  11"  X  1'  4"  X  10". 

Drainage  Pipes 

In  order  to  have  pipes  of  a  proper  size  to  carry  away  the 
water  from  the  roof  of  a  building,  it  is  necessary  to  know  the 
number  of  gallons  of  water  that  will  vbe  drained.  To  find  the 
number  of  gallons  that  will  drain  from  a  roof  in  a  month, 
multiply  the  number  of  square  feet  of  the  roof  by  the  average 
number  of  inches  of  rainfall  per  month,  and  multiply  this  prod- 
uct by  .623.  When  the  roof  is  not  flat,  or  very  nearly  so,  its 
area  should  be  considered  as  the  area  which  it  actually  covers. 


^  EXAMPLES 

11.  A  roof  is  48'  by  62'.     How  many  gallons  of  water  will 
drain  from  it  in  a  month  if  the  rainfall  is  6"  ? 

12.  A  roof  is  29'  by  74'.     How  many  gallons  of  water  will 
drain  from  it  if  the  rainfall  is  4^^"  ? 


PLUMBING  AND  HYDRAULICS  175 

Note. — The  rainfall  per  month  in  Massachusettjs  varies  from  J"  to 
12'^  In  calculating  the  following  problems  consider  10"  rainfall  as  the 
average  amount. 

13.  Find  the  number  of  gallons  that  will  drain  from  a  flat 
roof  (a)  112'  by  64';  (6)  88'  by  49';  (c)  120'  by  80'. 

14.  Find  the  number  of  gallons  that  will  drain  from  a  slant- 
ing roof  each  half  of  which  is  (a)  52'  by  34' ;  {b)  49'  by  28' ; 
(c)  112'  by  54';  (d)  57'  by  33';  (e)  54'  by  31'. 

Weight  of  Lead  Pipe 

Pipes,  particularly  those  of  lead,  are  sold  by  weight,  aiid  this 
depends  upon  the  diameter  and  thickness  of  the  metal  in  the 
pipe. 

To  find  the  weight  of  a  length  of  pipe,  subtract  the  square 
of  the  inner  diameter  in  inches  from  the  square  of  the  outer 
diameter  in  inches,  and  multiply  the  remainder  by  the  weight 
of  12  cylindrical  inches.  This  product  multiplied  by  the 
length  in  feet  gives  the  required  weight. 

Example.  —  What  is  the  weight  of  1450  feet  of  lead  pipe, 
the  outer  diameter  being  J"  and  the  inner  diameter  being  ^"? 
D  =  outer  diameter 
d  =  inner  diameter 
I  =  leiijith  of  pipe 
3.8698  lb.  =  weight  of  12  cylindrical  inches  of  lead  pipe  1'  long 
and  1"  in  diameter.     Weight  in  lb.  =  (Z>-  —  d-)  x  3.8698  x  I 

Then  ^^  =  LJLL  -  J^  (sciuare  of  outer  diameter) 

(8)2     8  X  8     64  ^   *  -         ^ 

VJ  «  =   ^  ^    ^  =  — ^  (square  of  inner  diameter) 
(16)2     16  X  16     256  ^  ^  ^ 

15  _  i5.  =  iL  =  .5742  (difference) 
64     266     256  ^  ^ 

.6742  X  3.8698  x  1450  =  3221.94  lb. 

EXAMPLES 

1.  What  is  the  weight  of  364'  of  lead  pipe,  outer  diameter 
}",  inner  diameter  ^y  ? 


176  VOCATIONAL  MATHEMATICS 

2.  What  is  the  weight  of  1189'  of  lead  pipe,  outer  diameter 
If",  inner  diameter  ly\"  ? 

3.  What  is  the  weight  of  2189'  of  lead  pipe,  outer  diameter 
2f' ',  inner  diameter  2^''  ? 

4.  What  is  the  weight  of  112'  of  lead  pipe,  outer  diameter 
3",  inner  diameter  2^%"  ? 

5.  What  is  the  weight  of  212'  of  lead  pipe,  outer  diameter 
3i",  inner  diameter  3^^"  ? 

Capacity  of  Pipe 

Rules  for  Finding  the  Capacity  in  Gallons  of  a  Foot  of  Pipe 
of  any  Diameter : 

1.  Find  the  cubical  contents  in  inches,  and  divide  by  231, 

C=D2x.7854x  12-- 231 

2.  Multiply  the  square  of  the  inside  diameter  by  .0408. 

C=D^X  .0408 

Rules  for  Findiyig  the  Capacity  of  a  Pipe  of  any  Length  and 
any  Diameter : 

1.  Multiply  the  number  of  gallons  in  a  foot  of  the  pipe  by 
the  number  of  feet  in  the  length  of  the  pipe. 

0=Z)2  X  .0408  X  L 

2.  Find  the  cubical  contents  in  inches,  and  divide  by  231. 

,(7=7>2  X  .7854  X  i.^231 

Note.  —  In  the  above  formulas,  let  C  =  the  capacity  of  the  pipe  in 
gallons,  D  =  the  diameter  in  inches,  and  L  =  the  length  in  feet. 

EXAMPLES 

1.  What  is  the  capacity  of  a  pipe  having  an  inside  diameter 
of  (a)  J  inch,  16  feet  in  length  ;  (b)  1}  inches,  20  feet  in  length  ; 
(c)  1.^  inches,  1  foot  in  length;  (d)  3 J  inches,  1  foot  in  length; 
(e)  14  inches,  10  feet  in  length  ;  (/)  5  inches^  22  feet  in  length  ? 


PLUMBING  AND  HYDRAULICS 


177 


Number  of  U.  S.  Gallons  in  Round  Tanks 
for  onb  foot  in  defth 


DiA.  or 
Tanks 
Ft.  1«. 

No. 
U.  S. 
Gals. 

CiTBic  Ft. 
AND  Area 
in  8q.  Ft 

DlA.  OF 

Tanks 
Ft.  In. 

No. 
U.S. 
Gals. 

Cub  in  Ft. 
ANo  Area 

IN  9«i.  Ft. 

DlA.  Of 

Tanks 
Fr.  In. 

No. 
U.S. 
Gals. 

Cdhio  Ft. 
ANi»  Arka 
iH  8m.  Ft. 

1 

5.87 

.7S5 

3 

52.88 

7.0<J9 

5 

146.88 

19.63 

1-1 

6.8» 

.922 

3-1 

55.86 

7.467 

6-1 

151.82 

20.29 

1-2 

8. 

1.069 

li-2 

68.92 

7.876 

5-2 

16(J.83 

20.97 

1-3 

9.18 

1.227 

3-3 

62.0(5 

8.296 

5-3 

161.93 

21.65 

1-4 

10.44 

1.396 

3-4 

65.28 

8.727 

6-4 

167.12 

22.34 

1-6 

11.79 

1.576 

3-6 

68.68 

9.168 

6-5 

172.38 

23.04 

1-6 

13.22 

1.767 

3-6 

71.97 

9.621 

5-6 

177.72 

23.76 

1-7 

14.73 

1.969 

3-7 

75.44 

10.085 

5-7 

183.15 

24.48 

l-« 

16.32 

2.182 

3-8 

78. 9i) 

10.569 

6-8 

188.66 

25.22 

1-9 

17.99 

2.405 

3-9 

82.62 

11.046 

6-9 

194.25 

25.97 

1-10 

19.75 

2.640 

3-10 

86.33 

11.541 

6-10 

199.92 

26.73 

l-ll 

21.68 

2.886 

3-11 

90.13 

12.048 

6-11 

205.67 

27.49 

2 

23.50 

3  142 

4 

94. 

12.666 

6 

211.51 

28.27 

2-1 

25.60 

3.409 

4-1 

97.96 

13.096 

6-3 

229.50 

30.68 

2-2 

27.68 

3.687 

4-2 

102. 

13.635 

6-6 

248.23 

33.18 

2-3 

29.74 

3.976 

4-3 

106.12 

14.186 

6-9 

267.69 

36.78 

2-4 

31.99 

4.276 

4-4 

110.32 

14.748 

7 

287  88 

38.48 

2-6 

34.31 

4.687 

4-6 

114.61 

15.321 

7-3 

308.81 

41.28 

2-6 

36.72 

4.909 

4-6 

118.97 

15.90 

7-6 

330.48 

44.18 

2-7 

39.21 

6.241 

4-7 

123.42 

16.50 

7-9 

352.88 

47.17 

2-8 

41.78 

5.585 

4-8 

127.95 

17.10 

8 

376.01 

50.27 

2-0 

44.43 

5.940 

4-9 

132.56 

17.72 

8-3 

399.88 

63.46 

2-10 

47.16 

6.305 

4-10 

137.25 

18.36 

8-6 

424.48 

66.75 

2-11 

40.98 

6.681 

4-11 

142  02 

18.99 

8-9 

449  82 

60.13 

31 J  gallons  equal  1  barrel. 

To  find  the  capacity  of  tanks  greater  than  the  largest  given  in  the  table, 
look  in  the  table  for  a  tank  one  half  of  the  given  size  and  multiply  it8 
capacity  by  4,  or  one  of  one  third  its  size  and  multiply  its  capacity 
by  9,  etc. 


178  VOCATIONAL  MATHEMATICS 

EXAMPLES 
Solve  the  following  problems  by  use  of  tables : 

1.  Find  the  number  of  gallons  contained  in  a  round  tank, 
1'  8"  in  diameter  and  V  deep. 

2.  Find  the  number  of  gallons  contained  in  a  round  tank, 
4'  11"  in  diameter  and  3'  deep. 

3.  Find  the  number  of  gallons  contained  in  a  round  tank, 
9'  3"  in  diameter  and  10'  deep. 

4.  Find  the  number  of  barrels  contained  in  a  round  tank, 
2'  4"  in  diameter  and  2'  deep. 

5.  Find  the  number  of  barrels  contained  in  a  round  tank, 
12'  6"  in  diameter  and  4'  deep. 

REVIEW  PROBLEMS 

1.  A  three-fourths-inch  pipe  has  an  actual  internal  diameter 
ot  .824  inch ;  what  is  the  nearest  rule  measure  to  a  j^g"  ? 

2.  A  two-inch  pipe  is  .154  inch  in  metal  thickness ;  what 
is  the  nearest  rule  measure  to  a  3^2"  '■* 

3.  A  one-inch  pipe  one  foot  long  holds  .373  pound  of  water ; 
how  long  should  it  be  to  hold  5  pounds  ? 

4.  A  five-inch  pipe  one  foot  long  holds  1.038  gallons;  how 
many  gallons  will  a  pipe  277  feet  long  hold  ? 

5.  If  a  five-inch  steam  pipe  has  an  actual  inside  diameter 
of  5.045  inches  and  an  actual  outside  diameter  of  5.565  inches, 
what  is  the  thickness  of  the  pipe  ? 

6.  The  inside  diameter  of  a  cylinder  is  22.625  inches  before 
boring  and  22.875  inches  after  boring ;  how  much  larger  is  the 
opening  after  boring  ? 

7.  What  must  be  the  height  of  a  rectangular  tank,  with 
base  12"  by  12",  to  hold  60  gallons  ? 

8.  What  is  the  height  of  a  rectangular  tank  with  a  base 
12"  X  14",  which  will  hold  189  gallons  ? 


PLUMBING  AND  HYDRAULICS  179 

9.    What  size  cylindrical  tank,  with  a  base  of  40  square 
Miches,  will  hold  89  gallons  ? 

10.  What  size  cylindrical  tank,  with  a  base  of  GO  square 
inches,  will  hold  42  gallons? 

U.  What  size  cylindrical  tank,  with  a  height  of  8  feet,  will 
hold  24  galloiis  ? 

12.  What  size  cylindrical  tank,  with  a  height  of  6  feet  4 
inches,  will  hold  62  gallons  ? 

13.  What  size  cylindrical  tank,  with  a  base  of  14  square 
inches,  will  hold  35  gallons  ? 

14.  What  size  rectangular  tank,  with  a  base  14  inches  X  7 
inches,  will  hold  14  gallons  ? 

Capacity  of  Pipes 

Law  of  Squares.  —  The  areas  of  similar  figures  vary  as  the 
squares  of  their  corresponding  dimensions. 

Pipes  are  cylindrical  in  shape  and  are,  therefore,  similar 
figures.  The  areas  of  any  two  pipes  are  to  each  other  as  the 
squares  of  the  diameters. 

Example.  —  If  one  pipe  is  4"  in  diameter  and  another  is  2" 
in  diameter,  their  ratio  is  J^,  and  the  area  of  the  larger  one  is, 
therefore,  4  times  the  smaller  one. 

EXAMPLES 

1.  How  much  larger  is  a  section  of  5"  pipe  than  a  section 
of  2"  pipe  ? 

2.  How  much  larger  is  a  section  of  2^"  pipe  than  a  section 
of  1"  pipe  ? 

3.  How  much  larger  is  a  section  of  5"  pipe  than  a  section  of 
3"  pipe  ? 

Rule  of  Thumb  Method.  —  This  method  is  used  for  finding 
the  size  of  a  pipe  necessary  to  fill  a  number  of  smaller  pipes, 
and  is  explained  as  follows : 


180 


VOCATIONAL  MATHEMATICS 


To  find  the  diameter  of  a  pipe  which  will  fill  three  pipes,  having  diameters 
of  2",  21",  and  4",  respectively.  Draw  a  right  angle,  one  arm  {AB)  be- 
ing 2"  in  length,  the  other  arm  (AC)  being  2^"  in  length.  Connect  the 
points  B  and  C.  The  length  of  this  line 
{BC)  will  give  the  size  of  a  pipe  neces- 
sary to  fill  the  two  smaller  pipes  (about 
S\").  From  one  end  of  this  last  line 
(BC)  and  at  right  angles  to  it  draw 
another  line  (BD)  4"  in  length.  Con- 
nect the  points  C  and  D.  The  length 
of  this  line  (CD)  will  represent  the  size  of  a  pipe  necessary  to  fill  the 
three  pipes,  having  diameters  of  2",  2J",  and  4",  respectively. 

This  process  may  be  continued  for  as  many  pipes  as  desired. 

The  carrying  capacity  of  circular  pipes  for  conveying  liquids 
depends  upon  the  area  of  the  opening  of  the  pipe.  Since  the 
areas  of  pipes  are  proportional  to  the  squares  of  the  diameters, 
the  principle  of  the  right  triangle  may  be  used  for  determining 
an  area  equivalent  to  two  or  more  areas,  or  a  pipe  equivalent 
to  three  or  more  pipes. 

Geometric  Proof.  —  (See  above  figure)  : 
In  the  rt.  A  ABC 

C^  =  AK  +  A(? 
In  the  rt.  A  CBD 


gd''=cb'  +  bd' 


Let 


CD  =  V(7J3'  +  BD" 
AB  =  2" 


^0=21" 


I      2  5M  _   4  1" 


BD  =  4" 
C&  =  4" 

CB=  V^7"  =  V"  =  ''^-2" 

Substituting  the  values  of  AB,  AC,  and  BC  in  the  above 
equations : 

OT' =(2)2+ (21)2  =  .V- 

CD' = -V  +  42  =  4^  +  16  =  1 J^" 

CD  =i^i^"  =  5.12" 


PLUMBING   AND   HYDRAULICS 


181 


EXAMPLES 

1.  Find  the  size  of  pipe  necessary  to  fill  a  1^",  2",  and  3" 
pipe. 

2.  Find  the  size  of  pipe  necessary  to  fill  a  J",  IJ",  and  2^' 
pipe. 

3.  Find  the  size  of  pipe  necessary  to  till  a  H",  -i",  and  3^" 
pipe. 

Atmospheric  Pressure 

Atmospheric  pressure  is  often  expressed  as  a 
certain  number  of  "  atmospheres."  The  pressure 
of  one  "atmosphere"  is  the  weight  of  a  column  of 
air,  one  square  inch  in  area. 

At  sea  level  the 
average  pressure  of 
the  atmosphere  is 
approximately  15 
pounds  per  square 
inch. 

The  pressure  of 
the  air  is  measured 
by  an  instrument 
called  a  barometer. 
The  barometer  con- 
sists of  a  glass  tube, 
about  31^  inches 
long,  which  has 
been  entirely  filled 
with  mercury  (thus 

removing  all   air  from  the  tube)  and  inverted  in 
a  vessel  of  mercury. 

The  space  at  the  top  of  the  column  of  mercury 
varies  as  the  air  pressure  on  the  surface  of  the 
mercury  in  the  vessel  increases  or  decreases.     The 
Bakohetkb  pressure  is  read  from  a  graduated  scale  which  indi- 


Barometer  Tube 


182  VOCATIONAL  MATHEMATICS 

cates  the   distance  from  the   surface  of  the  mercury  in  the 
vessel  to  th©  top  of  the  mercury  column  in  the  tube. 

QUESTIONS 

1.  Four  atmospheres  would  mean  how  many  pounds  ? 

2.  Give  in  pounds  the  following  pressures:  1  atmosphere; 
-^  atmosphere ;  J  atmosphere. 

3.  If  the  air,  on  the  average,  will  support  a  column  of 
mercury  30  inches  high  with  a  base  of  1  square  inch,  what 
is  the  pressure  of  the  air  ?  (One  cubic  foot  of  mercury  weighs 
849  pounds.) 

Water  Pressure 

When  water  is  stored  in  a  tank,  it  exerts  pressure  against 
the  sides,  whether  the  sides  are  vertical,  oblique,  or  horizontal. 
The  force  is  exerted  perpendicularly  to  the  surface  on  which  it 
acts.  In  other  words,  every  pound  of  water  in  a  tank,  at  a 
height  above  the  point  where  the  water  is  to  be  used,  possesses 
a  certain  amount  of  energy  due  to  its  position. 

It  is  often  necessary  to  estimate  the  energy  in  the  tank  at 
the  top  of  a  house  or  in  the  reservoir  of  a  town  or  city,  so  as 
to  secure  the  needed  water  pressure  for  use  in  case  of  fire.  In 
such  problems  one  must  know  the  perpendicular  height  from 
the  water  level  in  the  reservoir  to  the  point  of  discharge.  This 
perpendicular  height  is  called  the  head. 

Pressure  per  Square  Inch.  —  To  find  the  pressure  per  square 
inch  exerted  by  a  column  of  water,  multiply  the  head  of  water 
in  feet  by  0.434.     The  result  will  be  the  pressure  in  pounds. 

The  pressure  per  square  inch  is  due  to  the  weight  of  a  cohunn  of 
water  1  square  inch  in  area  and  the  height  of  the  cohimn.  Therefore, 
the  pressure,  or  weight  per  square  inch,  is  equal  to  the  weight  of  a  foot  of 
water  with  a  base  of  1  square  inch  multiplied  by  the  height  in  feet.  Since 
the  weight  of- a  column  of  water  1  foot  high  and  having  a  base  of  1  square 
inch  is  0.434  lb.,  we  obtain  the  pressure  per  square  inch  by  multiply- 
ing the  height  in  feet  by  0.434. 


PLUMBING  AND  HYDRAULICS 


183 


EXAMPLES 

What  is  the  pressure  per  square  inch  of  a  column  of  water 
(a)  8'  high?  (b)  15' 8"  high?  (c)  30' 4"  high?  (d)  18' 9" 
high?    (e)  41' 3"  high? 

Head.  —  To  find  the  head  of  water  in  feet,  if  the  pressure 
(weight)  per  square  inch  is  known,  multiply  the  pressure  by 
2.31. 

Let  p  =  pressure 

h  =  height  in  feet 
Then  2y=hx  0.434  lb.  per  sq.  in. 

0.434      0.434      ^  ^ 


EXAMPLES 

Find  the  head  of  water,  if  the  pressure  is  (a)  49  lb.  per 
sq.  in.;  (6)  88  lb.  per  sq.  in.;  (c)  46  lb.  per  sq.  in.;  (d) 
28  lb.  per  sq.  in. ;  (e)  64  lb.  per  sq.  in. 

Lateral  Pressure.  — To  find  the  lateral 
(sideways)  pressure  of  water  upon  the 
sides  of  a  tank,  multiply  the  area  of  the 
submerged  side,  in  inches,  by  the  pressure 
due  to  one  half  the  depth. 

Example.  — A  tank  18"  long  and  12" 
deep  is  full  of  water.  What  is  the  lateral 
pressure  on  one  side  ? 


lenirth  depth 

18"    X    12" 

depth 

1'       X  0.434 


216  square  inches  =  area  of  side 


.434  lb.  pressure  at  the  bottom  of 
the  tank 
0  =  pressure  at  top 
2) .434  lb. 

.217  lb.  average  pressure  due  to  one  half  the 

depth  of  the  tank 
.217  X  216  =  46.872  pounds  =  pressure  on  one 
side  of  the  tank 


U  sero 


U  half  (hat  at 


Lateral  Pbessubs 


184  VOCATIONAL  MATHEMATICS 

EXAMPLES 

1.  Find  the  lateral  pressure  on  one  end  of  a  tank  (a)  11" 
long,  8"  wide,  and  7"  deep;  (b)  4'  long,  5'  wide,  and  16"  deep. 

2.  Find  the  lateral  pressure  on  one  side  of  a  tank  (a)  8" 
long,  4"  wide,  and  6"  deep;  (b)  2'  long,  3'  wide,  and  11"  deep. 

3.  Find  the  total  pressure  on  the  bottom  of  a  tank  7'  long, 
3'  wide,  and  11"  deep. 

4.  Find  the  force  acting  on  the  bottom  of  a  box  9'  long,  6' 
wide,  and  2'  deep,  filled  with  water. 

Thickness  of  Pipe.  —  To  find  the  thickness  of  a  lead  pipe 
necessary  for  a  given  head  of  water,  multiply  the  head  in  feet 
by  the  size  of  the  pipe  required,  expressed  as  a  decimal,  and 
divide  this  result  by  750.  The  quotient  will  represent  the 
thickness  required  in  hundredths  of  an  inch. 

h  X  .s 


T= 


750 


T  =  thickness  of  pipe 

.s  =  size  of  pipe  expressed  as  a  decimal  of  an  inch 

h  =  head  of  water 

Example.  —  What  thickness  should  a  half-inch  pipe  have 
for  a  50-foot  head  of  water  ? 

50  X  .5  =  25  25  -  750  =  .03  inch 

EXAMPLES 

1.  What  thickness  of  ^"  pipe  is  necessary  for  a  head  of  28'? 

2.  What  thickness  of  1"  pipe  is  necessary  for  a  head  of  64'? 

3.  What  thickness  of  1^"  pipe  is  necessary  for  a  head  of 
112'? 

4.  What  thickness   of  2"  pipe  is  necessary  for  a  head  of 
250'? 

5.  What  thickness  of  2i"  pipe  is  necessary  for  a  head  of 
275'? 


PLUMBING   AND  HYDRAULICS 


185 


Water  Traps 

The  question  of  disposing'  of  the  waste  water,  called  sewage,  is  of 

leat  importance.     Various  devices  may  be  used  to  prevent  odors  from 

the  sewage  enterin<;  the  house.     In  order  to  prevent  the  escape  of  gas 


Watkr  Traps 

from  the  outlet  of  the  sewer  in  the  basement  of  a  house  or  building,  a 
device  called  a  trap  is  used.  This  trap  consists  of  a  vessel  of  water 
placed  in  the  waste  pipe  of  the  plumbing  fixtures.  It  allows  the  free  pas- 
sage of  waste  material,  but  prevents  sewer  gases  or  foul  odors  from  enter- 
ing the  living  rooms.  The  vessels  holding  the  water  have  different  forms  ; 
(see  illustration).  These  traps  may  be  emptied  by  back  pressure  or  by 
siphon.  It  is  a  good  plan  to  have  sufficient  water  in  the  trap  so  that  it 
will  never  be  empty.  All  these  problems  belong  to  the  plumber  and  in- 
volve more  or  less  arithmetic. 


To  determine  the  pressure  which  the  seal  of  a  trap  will  resist : 
Example.  —  What  pressure  will  a  l^inch  trap  resist  ? 

If  one  arm  of  the  trap  has  a  seal  of  1 J  inches,  both  arms  will  make  a 
column  twice  as  high,  or  3  inches.  Since  a  column  of  water  28  inches 
in  height  is  equivalent  to  a  pressure  of  1  pound,  or  16  ounces,  a  column 
of  water  1  inch  in  height  is  equivalent  to  a  pressure  of  ^^  of  a  pound,  or 
if  =  f  ounces,  and  a  column  of  water  3  inches  in  height  is  equivalent  to 
3  X  f  =  J^  =  1.7  ounces.       • 

Therefore,  a  IJ-inch  trap  will  resist  1.7  ounces  of  pressure. 


186  VOCATIONAL  MATHEMATICS 

EXAMPLES 

1.  What  back  pressure  will  a  f-inch  seal  trap  resist  ? 

2.  What  back  pressure  will  a  2-inch  seal  trap  resist  ? 
.3.   What  back  pressure  will  a  2^inch  seal  trap  resist  ? 

4.  What  back  pressure  will  a  4^inch  seal  trap  resist  ? 

5.  What  back  pressure  will  a  5-inch  seal  trap  resist  ? 

Velocity  of  Water 

Velocity  through  Pipes.  —  To  calculate  the  velocity  of  water 
flowing  through  a  horizontal  straight  pipe  of  given  length  and 
diameter,  the  head  of  water  above  the  center  of  the  pipe  being 
known,  multiply  the  head  of  water  in  feet  by  2500  and  divide 
the  result  by  the  length  of  the  pipe  in  feet  multiplied  by  13.9, 
divided  by  the  inner  diameter  of  the  pipe  in  inches.  The 
square  root  of  the  quotient  gives  the  velocity  in  feet  per 
second. 

Let       I  =  length  of  the  pipe  in  feet 
d  =  diameter  of  the  pipe 
H=  head  of  water  above  the  center  of  pipe 
F=  velocity  of  water  in  feet  per  second 


Then  V=y\E^Ll^ 
d 


Example.  —  The  head  of  water  is  6  feet,  the  length  of  the 
pipe  1340  feet,  and  its  diameter  5  inches.  What  is  the  velocity 
of  water  passing  through  the  pipe  ? 

Substituting  in  the  above  formula  : 


=v 


6  X  2600  / 15000  r-—      „  ^ 


PLUMBING   AND   HYDRAULICS  187 

EXAMPLES 

What  is  the  velocity  of  water  running  through  the  following 
pipes: 

1.  Head  16';  length  of  pipe,  811';  diameter  of  pipe,  6"  ? 

2.  Head  10' ;  length  of  pipe,  647' ;  diameter  of  pipe,  4"  ? 

3.  Head  14';  length  of  pipe,  489';  diameter  of  pipe,  4"  ? 

4.  Head  26';  length  of  pipe,  264';  diameter  of  pipe,  2^"? 
Head  and  Velocity.  —  To  find  the  head  which  will  produce  a 

given  velocity  of  water  through  a  pipe  of  a  given  diameter  and 
length,  multiply  the  square  of  the  velocity,  expressed  in  feet 
per  second,  by  the  length  of  pipe,  multiplied  by  the  quotient 
obtained  by  dividing  13.9  by  the  diameter  of  the  pipe  in  inches. 
Divide  this  result  by  2500,  and  the  final  quotient  will  give  the 
head  in  feet. 

13.9 


VxLx 


d 


2500 

F=  velocity  of  water  expressed  in  ft.  per  sec. 

L  =  length  of  pipe  expressed  in  feet 

d  =  diameter  of  pipe  expressed  in  inches 

Example. —  The  horizontal  length  of  pipe  is  1200  feet  and 
the  diameter  is  4  inches.  What  head  must  be  secured  to 
produce  a  flow  of  3  feet  per  second  ? 

T^xLxlM  3 

d        9xZg00xl3.9      27x13  9 

2500         -  ix'^m    ~        26        -^^'    ^'**- 

1        25 

EXAMPLES 

1.  The  horizontal  length  of  a  pipe  is  845  feet  and  the  diam- 
eter is  3  feet.  What  head  must  be  secured  to  produce  a  flow 
of  2\  feet  per  second  ? 

2.  The  horizontal  length  of  a  pipe  is  980  feet  and  the  diam- 
eter is  2^  inches.  What  head  must  be  secured  to  produce  a 
flow  of  4  feet  per  second  ? 


188  VOCATIONAL  MATHEMATICS 

3.  The  horizontal  length  of  a  pipe  is  1500  feet  and  the  diam- 
eter is  3  inches.  What  head  must  be  secured  to  produce  a 
flow  of  4  feet  per  second  ? 

4.  The  horizontal  length  of  a  pipe^  is  1280  feet  and  the 
diameter  is  2i  inches.  AVhat  head  must  be  secured  to  pro- 
duce a  flow  of  S^  feet  per  second  ? 

5.  The  horizontal  length  of  a  pipe  is  1890  feet  and  the 
diameter  is  4  inches.  What  head  of  water  must  be  secured 
to  produce  a  flow  of  5  feet  per  second  ? 

Power 

Power  is  the  time  rate  of  doing  work.  The  unit  of  power  is 
the  horse  power  (H.  P.),  which  represents  33,000  foot  pounds  a 
minute  or  550  foot  pounds  a  second.  A  foot  pound  is  the 
amount  of  work  necessary  to  lift  a  pound  through  a  distance 
of  one  foot. 

Raising  Water.  —  To  find  the  power,  necessary  to  raise  water 
to  any  given  height,  multiply  the  product  of  cubic  feet  required 
per  minute  and  the  number  of  feet  through  which  it  is  to  be 
lifted  by  62.3  and  divide  this  product  by  33,000.  This  will 
give  the  nominal  horse  power  required.  If  the  amount  of 
water  required  per  minute  is  in  gallons,  the  multiplier  should 
be  8.3  instead  of  62.3. 

EXAMPLES 

1.  Find  the  power  necessary  to  raise  a  bucket  (weighing 
2  lb.  when  empty)  holding  a  quart  of  water  24'  in  4  seconds. 

2.  Find  the  power  necessary  to  raise  3  cubic  feet  of  water 
15'  in  30  seconds. 

3.  Find  the  power  necessary  to  raise  5  cubic  feet  of  water 
29'  in  44  seconds. 

4.  Find  the  power  necessary  to  raise  1  gallon  of  water  31' 
in  2  seconds. 

Note.  — Change  quarts  to  fractions  of  a  gallon. 


PLUMBING    AND    HYDRAULICS  189 

Water  Power 

When  water  flows  from  one  level  to  another,  it  exerts  a 
certain  amount  of  energy,  which  is  the  capacity  for  doing 
work.  This  energy  may  be  utilized  by  such  means  as  the 
water  wheel,  the  turbine,  and  the  hydraulic  ram. 

Friction,  which  must  be  considered  when  one  speaks  of 
water  power,  is  the  resistiuice  which  a  substance  encounters 
when  moving  through  or  over  another  substance.  The  amount 
of  friction  depends  upon  the  pressure  between  the  surfaces  in 
contact. 

When  work  is  done  a  part  of  the  energy  which  is  expended 
is  apparently  lost.  In  the  case  of  water  this  is  due  to  the 
friction.  All  the  energy  which  the  water  has  cannot  be  used 
to  advantage,  and  efficiency  is  the  ratio  of  the  useful  work  done 
by  the  water  to  the  total  work  done  by  it. 

Efficiency.  —  To  find  the  amomit  of  useful  work  done  when 
a  pump  lifts  or  forces  water  to  a  higher  level,  multiply  the 
weight  of  the  water  in  pounds  by  the  height  in  feet  through 
which  it  is  raised. 

Since  friction  must  be  taken  into  consideration,  the  total 
work  done  by  the  pump  will  be  greater  than  the  useful  work. 
To  find  the  amount  of  the  total  work,  multiply  the  amount  of 
the  useful  work  by  the  reciprocal  of  the  efliciency  of  the  pump. 

Example.  —  Find  the  power  required  to  raise  half  a  ton 
(long  ton,  or  2240  lb.)  of  water  to  a  height  of  40  feet  in  a 
minute,  when  the  efficiency  of  the  pump  is  75  %. 

Total  work  done  =  weight  x  height  x  efficiency  counter. 
1120  X  40  X  -V^  =  59,783  ft.  lb. 

H.  P.  required  =  ^^j1^  =  1.8.     Ans. 
*  33,000 

EXAMPLES 

1.  Find  the  power  required  to  raise  a  cubic  foot  of  water 
28'  in  8  miimtes,  if  the  pump  has  80  %  efficiency. 


190  VOCATIONAL    MATHEMATICS 

2,  Find  the  power  required  to  raise  a  gallon  of  water  16' 
in  6  seconds,  if  the  pump  has  85  %  efficiency. 

3.  Find  the  power  required  to  raise  a  quart  of  water  26'  in 
12  seconds,  if  the  pump  has  70  %  efficiency. 

Cement  and  Solder 

Cement.  —  In  making  joints  on  earthenware  house  sewers, 
plumbers  use  a  mixture  of  one  part  Portland  cement,  and  two 
parts  clean  sand,  over  a  ring  of  oakum. 

Portland  cement  is  hydraulic  lime  and  is  made  by  burning  a  mixture 
of  limestone,  clay,  and  sand,  and  grinding  the  product  to  a  very  fine 
powder.  Oakum  is  made  from  old  ropes,  which  have  been  picked  to 
pieces.  After  these  ropes  have  been  untwisted,  loosened,  and  drawn 
out,  the  material,  oakum,  is  lieated  with  tar  to  make  it  flexible. 

Solder. — In  plumbing  it  is  often  desirable  to  join  two 
metals  or  two  pieces  of  the  same  metal.  This  may  be  done  by 
means  of  an  alloy  called  solder,  which  melts  at  a  lower  tem- 
perature than  the  metals  that  are  to  be  united.  This  alloy 
melts  at  a  comparatively  low  temperature  (below  500°  F.)  and 
will  hold  the  two  pieces  together  unless  they  are  subject  to 
great  heat  or  stress.  The  process  of  uniting  metal  surfaces  in 
this  way  is  called  soldering. 

Solder  is  made  of  tin  and  lead  dross  mixed  with  resin  and  charcoal, 
and  heated  in  a  furnace  covered  with  a  hood.  After  melting  and 
stirring,  the  product  is  drawn  off  into  a  small  pot  furnace,  dipped  out 
with  ladles,  and  run  into  molds. 

There  are  different  grades  of  solder:  soft,  hard,  etc.  Soft  solder  con- 
sists of  two  parts  of  tin  and  one  part  of  lead,  and  melts  at  about  340°  F. 
The  addition  of  bismuth  makes  it  more  fusible.  Hard  solder,  composed 
of  equal  parts  of  tin  and  lead,  is  used  by  the  tinsmith. 

A  solder  used  by  plumbers  for  wiping  is  composed  of  three  parts  of 
lead  and  two  parts  of  tin. 

Hard  spelter  is  made  of  one  part  of  copper  and  one  part  of  zinc,  while 
soft  spelter  is  made  from  two  parts  of  copper  and  three  parts  of  zinc. 

Solder  is  applied  by  means  of  a  soldering  iron  made  of  copper  and 


PLUMBING  AND  HYDRAULICS  191 

pointed  at  the  end.  The  handle  ia  made  of  wood  to  prevent  the  heat 
from  passing  from  tl»e  hot  iron  to  the  hand. 

To  solder  :  Brighten  the  copper  with  a  file,  moisten  with  acid,  and 
then  apply  the  solder  until  cooled. 

For  jointing,  practical  plumbers  allow  for  each  joint  one  inch  in  size  one 
pound  of  calking  lead. 

Example  — Estimate  the  pounds  of  calking  required  for 
18  4"  joints  a'nd  25  2"  joints. 

I  4"  joint  requires  4  lb.  lead  (calking) 
1  2"  joint  requires  2  lb.  lead  (calking) 
Therefore  18  4"  joints  require    72  lb. 
26  2"  joints  require    50  lb. 
122  lb. 

EXAMPLES 

1.  How  much  copper  and  zinc  will  be  used  (a)  to  make 
29  lb.  of  soft  spelter  ?     (6)  to  make  13  lb.  of  hard  spelter  ? 

2.  How  much  calking  lead  should  be  allowed  for  (a)  24 
8"  joints  and  18  2"  joints?  (b)  17  2^'  joints  and  21  4" 
joints  ?     (c)  41  2"  joints  and  38  2^"  joints  ? 

Density  of  Water 

The  mass  of  a  unit  volume  of  a  subtance  is  called  its  density. 

One  cubic  foot  of  pure  water  at  39.1°  F.  has  a  mass  of 
62.425  pounds ;  therefore,  its  density  at  this  temperature  is 
62.425,  or  approximately  62.5.  At  this  temperature  water  has 
its  greatest  density,  and  with  a  change  of  temperature  the 
density  is  also  changed. 

With  a  rise  of  temperature,  the  density  decreases  until  at 
212°  F.,  the  boiling  point  of  water,  the  weight  of  a  cubic  foot 
of  fresh  water  is  only  69.64  pounds. 

When  the  temperature  falls  below  39.1°  ¥._,  the  density  of 
water  decreases  until  we  find  the  weight  of  a  cubic  foot  of  ice 
to  be  but  57.5  pounds. 


192  VOCATIONAL  MATHEMATICS 

EXAMPLES 

1.  One  cubic  foot  of  fresh  water  at  62.5°  F.  weighs  62.355  lb., 
or  approximately  62.4  lb.  What  is  the  weight  of  1  cubic  inch  ? 
What  is  the  weight  of  1  gallon  (231  cubic  inches)  ? 

2.  What  is  the  weight  of  a  gallon  of  water  at  39.1°  F.  ? 

3.  What  is  the  weight  of  a  gallon  of  water  at  212°  F.  ? 

4.  What  is  the  weight  of  a  volume  of  ice  represented  by  a 
gallon  of  water  ? 

5.  What  is  the  volume  of  a  pound  of  water  at  ordinary 
temperature,  62.5°  F.  ? 

Specific  Gravity- 
Some  forms  of  matter  are  heavier  than  others,  i.e.  lead  is 
heavier  than  wood.  It  is  often  desirable  to  compare  the 
weights  of  different  forms  of  matter  and,  in  order  to  do  this, 
a  common  unit  of  comparison  must  be  selected.  Water  is 
taken  as  the  standard. 

Specific  Gravity  is  the  ratio  of  the  mass  of  any  volume  of  a 
substance  to  the  mass  of  the  same  volume  of  pure  water  at 
4°  C.  or  39.1°  F.  It  is  found  by  dividing  the  weight  of  a  known 
volume  of  a  substance  in  liquid  by  the  weight  of  an  equal 
volume  of  water. 

Example.  —  A  cubic  foot  of  wrought  iron  weighs  about 
480  pounds.     Find  its  specific  gravity. 

Note,  —  1  ou.  ft.  of  water  weighs  62.425  lb. 

Weight  of  1  cu.  ft.  of  iron    _     480     _  -  ^        .^ 
Weight  of  1  cu.  ft.  of  water      62.425 

To  find  Specific  Gravity. — To  find  the  specific  gravity  of  a 
solid,  weigh  it  in  air  and  then  in  water.  Find  the  difference 
between  its  weight  in  air  and  its  weight  in  water,  which  will 
be  the  buoyant  force  on  the  body,  or  the  weight  of  an  equal 
volume  of  water.  Divide  the  weight  of  the  solid  in  air  by  its 
buoyant  force,  or  the  weight  of  an  equal  volume  of  water,  and 
the  quotient  will  be  the  specific  gravity  of  the  solid. 


PLUMBING   AND   HYDRAULICS  193 

Tables  have  been  compiled  giving  the  specific  gravity  of  different  solids, 
so  it  is  seldom  necessary  to  compute  it. 

The  specific  gravity  of  liquids  is  very  often  used  in  the 
industrial  world,  as  it  means  the  "  strength  "  of  a  liquid.  In 
the  carbonization  of  raw  wool,  the  wool  must  be  soaked  in 
sulj)huric  acid  of  a  certain  strength.  This  acid  cannot  be 
bought  except  in  ^ts  concentrated  form  (sp.  gr.  1.84),  and  it 
must  be  diluted  with  water  until  it  is  of  the  required  strength. 

The  simplest  way  to  determine  the  specific  gravity  of  a  liquid  is  with 
a  hydrometer.  This  instrument  consists  of  a  closed  glass  tube,  with  a 
bulb  at  the  lower  end  filled  with  mercury.  This  bulb  of  mer- 
cury keeps  the  hydrometer  upright  when  it  is  imniersed  in  a 
liquid.  The  hydrometer  has  a  scale  on  the  tube  which  can 
be  read  when  the  instrument  is  placed  in  a  graduate  of  the 
liquid  whose  specific  gravity  is  to  be  determined. 

But  not  all  instruments  have  the  specific  gravity  recorded 
on  the  stem.  Those  most  commonly  in  use  are  graduated 
with  an  impartial  scale. 

In  England,  Twaddell's  scale  is  commonly  employed,  and 
since  most  of  the  textile  mill  workers  are  P^nglish,  we  find  the 
same  scale  in  use  in  this  country.  The  Twaddell  scale  bears  a 
marked  relation  to  specific  gravity  and  can  be  easily  converted 
into  it. 

Another  scale  of  the  hydrometer  is  the  Beaume,  but  these 
readings  cannot  be  converted  into  specific  gravity  without 
the  use  of  a  complicated  formula  or  reference  to  a  table.  ^ ^ 

nVPROMETKR    SCALK  FoBMlTLA    KOIC   CoWEKTIN*;    INTO    S.  G. 

1.  Specific  gravity  hydrometer  Gives  direct  reading 

2.  Twaddell  S.  G.  =  (Ax^UliOO 

100 

3.  Beaume  S.  G.  =      ^^"^ 


146.3  -  JV 
iV=  the  particular  degree  which  is  to  be  converted. 

ExAMPLK.  —  Change  108  degrees  (Tw.)  into  S.  G. 

(.5x168) +  100^,8^      ^,„ 
100  . 


194  VOCATIONAL  MATHEMATICS 

Another  formula  for  changing  degrees  Twaddell  scale  into  specific 

gravity  is  :  ^ ^-^ =  specific  gravity. 

1000 

In  Twaddell's  scale,  1°  specific  gravity  =  1.005 
2°  specific  gravity  =  1.010 
3°  specific  gravity  =  1.015 

and  so  on  by  a  regular  increase  of  .005  for  each  degree. 

To  find  the  degrees  Twaddell  when  the  specific  gravity  is  given,  multi- 
ply the  specific  gravity  by  1000,  subtract  1000,  and  divide  by  5.    Formula  : 

(S.G.X  1000) -1000  ^  ^  ^^^,^^11 

5  " 

Example.  —  Change  1.84  specific  gravity  into  degrees  Tv^^ad- 
dell, 

(1.84x1000) -1000  ^  igg  ^  T^^^^^ii 


EXAMPLES 

1.  What  is  the  specific  gravity  of  sulphuric  acid  of  116°  Tw.  ? 

2.  What  is  the  specific  gravity  of  acetic  acid  of  86°  Tw.  ? 

3.  What  is  the  specific  gravity  of  a  liquid  of  134°  Be.  ? 

4.  What  is  the  specific  gravity  of  a  liquid  of  108°  Be.  ? 

5.  What  is  the  specific  gravity  of  a  liquid  of  142°  Tw.  ? 

6.  Change  the  following  specific  gravity  readings  into  Tw : 

(a)  1.81;   (b)  1.12;  (c)  1.60;  (d)  1.44;  (e)  1.29. 


PART   Vn  — STEAM   ENGINEERING 

CHAPTER   XII 
HEAT 

Heat  Units. — The  unit  of  heat  used  in  the  industries  and 
shops  of  America  and  England  is  the  British  Thermal  Unit 
(B.  T.  U.)  and  is  defined  as  the  quantity  of  heat  required  to 
raise  one  pound  of  water  through  a  temperature  of  one  degree 
Fahrenheit. 

Thus  the  heat  required  to  raise  5  lb.  of  water  through  15 
degrees  F.  equals 

6  X  15  =  75  British  Thermal  Units  (B.  T.  U.) 
Similarly,  to  raise  86  lb.  of  water  through  ^°  F.  requires 
86  X  i  =  43  B.  T.  U. 

The  unit  that  is  used  on  the  Continent  and  among  scientific 
circles  in  America  is  the  metric  system  unit,  a  calorie.  This 
is  the  amount  of  heat  necessary  to  raise  1  gram  of  water 
1  degree  Centigrade.     (See  Appendix  for  metric  system.) 

EXAMPLES 

1.  How  many  units  (B.  T.  U.)  will  be  required  to  raise 
4823  lb.  of  water  62  degrees  ? 

2.  How  many  B.  T.  U.  of  heat  are  required  to  change  365 
cubic  feet  of  water  from  66"  F.  to  208**  F.  ? 

3.  How  many  units  (B.  T.  U.)  will  be  required  to  raise  785 
lb.  of  water  from  74°  F.  to  208°  F.  ? 

(Consider  one  cubic  foot  of  water  equal  to  62^  lb.) 

4.  How  many  B.  T.  U.  of  heat  are  required  to  change  1825 
cu.  ft.  of  water  from  118°  to  211°  ? 

106 


196 


VOCATIONAL  MATHEMATICS 


5.  How  many  heat  units  will  it  take  to  raise  484  gallons  of 
water  12  degrees  ? 

6.  How  many  heat  units  will  it  take  to  raise  5116  gallons 
of  water  from  66°  F.  to  198°  F.  ? 


212' 


Temperature 

The  ordinary  instruments  used  to  measure  temperature 
are  called  thermometers.  There  are  two  kinds  —  Fahren- 
heit and  Centigrade.  The  Fahrenheit  ther- 
mometer consists  of  a  cylindrical  tube  filled 
with  mercury  with  the  position  of  the  mercury 
at  the  boiling  point  of  water  marked  212,  and  the 
position  of  mercury  at  the  freezing  point  of 
water  32.  The  intervening  space  is  divided  into 
180  divisions.  The  Centigrade  thermometer  has 
the  position  of  the  boiling  point  of  water  100 
and  the  freezing  point  0.  The  intervening  space 
is  divided  into  100  spaces.  It  is  often  necessary 
to  convert  the  Centigrade  scale  into  the  Fah- 
renheit scale,  and  Fahrenheit  into  Centigrade. 

To  convert  F.  into  C,  subtract  32  from  the  F. 
degrees  and  multiply  by  -|  or  divide  by  1.8,  or 
C.  =  (F.  -  32°)  I,  where  C.  =  Centigrade  reading 
and  F.  =  Fahrenheit  reading. 

To  convert  C.  to  F.,  multiply  C.  degrees  by  f  or  1.8  and  add 
32. 


1 

D| 

100° 

..... 

g 

.y 

"S 

' 

a> 

, 

O 

r 

0° 

— 

"" 

—  17.8° 

c 

1 

II 


Thermometers 


90 


32° 


Example.  —  Convert  212  degrees  F.  to  C.  reading. 

5f212°-32°)  ^  5(180°)  ^  900^  ^  j^^o  q 
9  9  9' 

Example.  —  Convert  100  degrees  C.  to  F.  reading. 

^  ^  ^^^°  +  32°  =  ^^  +  32°  =  180°  +  32°  =  212°  F. 
6  6 


HEAT  197 

If  the  teinperatiire  is  below  the  freeziuj^  point,  it  is  usually 
written  with  a  minus  sign  before  it :  thus,  15  degrees  below  the 
freezing  point  is  written  —  15**.  In  changing  —  15°  ( -.  into  F. 
we  must  bear  in  mind  the  minus  sign. 

Thus,       F  =—  +  32^.     F  =  ~^'^°  ^  ^  +  32"  =  -  27-^  +  32°  =  6*^' 
o  5 

Example.  —  Change  -  22°  F.  to  C. 

C.  =  §  (F.  -  32°) 

C.  =  I  (-  22° -  82°)  =  |( -  64°)  =  -  30° 

EXAMPLES 

1.  Change  36°  F.  to  C.  6.  Change  225°  C.  to  F. 

2.  Change  89°  F.  to  C.  7.  Change  380°  C.  to  F. 

3.  Change  289°  F.  to  C.  8.  Change  415°  C.  to  F. 

4.  Change  350°  F.  to  C.  9.  Change  580°  C.  to  F. 

5.  Change  119°  C.  to  F. 

Value  of  Coal  in  Producing  Heat        ^ 

There  "are  different  kinds  of  coal  on  the  market.  Some 
grades  of  the  same  coal  give  off  more  heat  than  others  in 
burning.  The  heating  value  of  a  coal  may  be  found  in 
three  ways :  (1)  By  chemical  analysis  to  determine  the  amount 
of  carbon,  (2)  by  burning  a  definite  amount  in  a  calorimeter 
and  noting  the  rise  of  temperature  of  the  water,  (3)  by  actual 
trial  in  a  steam  boiler.  The  first  two  methods  give  a  theoreti- 
cal value,  the  third  gives  the  real  result  under  the  actual  con- 
ditions of  draft,  heating  surface,  combustion,  etc. 

EXAMPLES 

1.  If  125  lb.  of  ashes  are  produced  from  one  ton  of  coal, 
what  is  the  percentage  of  ashes  in  that  coal  ? 


198  VOCATIONAL  MATHEMATICS 

2.  Twelve  tons  of  coal  are  burned  per  day,  and  twenty -two 
baskets  of  ashes,  each  weighing  65  lb.,  are  removed;  what  per- 
centage of  ashes  does  the  coal  contain  ? 

3.  If  twelve  tons  of  coal  are  burned  per  day,  and  1450  lb.  of 
ashes  are  produced,  what  percentage  of  ashes  does  the  coal 
contain  ? 

4.  A  quantity  of  coal  is  built  into  a  rectangular  stack  50  ft. 
long,  25  ft.  broad,  and  6  ft.  high ;  what  is  the  weight  of  the 
coal,  allowing  45  cubic  feet  per  ton  ? 

5.  If  92,400  lb.  of  coal  are  consumed  in  60  hours  and  the 
engines  regularly  develop  480 1.  H.  P.  (Indicated  Horse  Power), 
how  much  coal  is  consumed  per  H.  P.  per  hour  ? 

6.  With  the  price  of  coal  at  $3.25  per  ton  of  2000  lb.,  and 
the  power  produced  earning  a  profit  of  25  %  on  the  cost  of 
production,  what  would  be  the  amount  of  profit  in  Ex.  5  when 
running  full  power  ? 

Latent  Heat 

By  latent  heat  of  water  is  meant  that  heat  which  water  ab- 
sorbs or  discharges  in  passing  from  the  liquid  to  the  gaseous, 
or  liquid  to  solid  state,  without  affecting  its  own  temperature. 
Thus,  the  temperature  of  boiling  water  at  atmospheric  pressure 
never  rises  above  212  degrees  F.,  because  the  steam  absorbs  the 
excess  of  heat  which  is  necessary  for  its  gaseous  state.  Latent 
heat  of  steam  is  the  quantity  of  heat  necessary  to  convert  a 
pound  of  water  into  steam  of  the  same  temperature  as  the 
steam  in  question. 

To  find  the  latent  heat  of  steam,  subtract  ten  times  the 
square  root  of  the  gauge  pressure  from  977. 

Example.  —  Find  the  latent  heat  of  steam  at  169  lb.  gauge 
pressure. 

Vim  =  13 
13  X  10  =  130 
977  -  180  =  847  B.  T.  U. 


HEAT  199 

EXAMPLES 

Find  the  latent  heat  of  steam  at  the  following  gauge 
pressures : 

1.  132  lb.  gauge  pressure.  6.  39  lb.  gauge  pressure. 

2.  116  lb.  gauge  pressure.  7.  41  lb.  gauge  pressure. 

3.  208  lb.  gauge  pressure.  8.  160  lb.  gauge  pressure. 

4.  196  lb.  gauge  pressure.  9.  159  lb.  gauge  pressure. 

5.  84  lb.  gauge  pressure.  10.  180J^  lb.  gauge  pressure. 

Heat  Units  Required  to  Produce  a  Given  Pressure 

To  lind  the  number  of  units  of  heat  required  to  raise  the 
temperature  corresponding  to  one  gauge  pressure  to  that  of 
another,  find  the  square  roots  of  the  gauge  pressures,  subtract 
these  values,  and  multiply  by  14J. 

Example.  —  Find  the  number  of  heat  units  required  to  raise 
the  temperature  of  64  pounds  gauge  pressure  to  169  pounds 
gauge  pressure. 

\/(l4  =  8  13-8  =  6         14^x6  =  711 

Vl«9  =  13  Approximately  72  B.  T.  U.     Am. 

EXAMPLES 

Find  the  number  of  heat  units  required  to  raise  the  tempera- 
ture between  the  following  gauges: 

1.  From  64  to  128  lb.  gauge.  5.  From  42  to  121  lb.  gauge. 

2.  From  26  to  131  lb.  gauge.  6.  From  28  to  132  lb.  gauge. 

3.  From  39  to  149  lb.  gauge.  7.  From  33  to  144  lb.  gauge. 

4.  From  49  to  165  lb.  gauge.  8.  From  55  to  164  lb.  gauge. 

Volume  of  Water  and  Steam  » 

According  to  steam  tables  one  cubic  foot  of  steam  at  100 
pounds'  pressure  weighs  0.2307  lb.,  one  cubic  foot  of  water 
weighs  62^  lb.,  and  one  gallon  of  water  may  be  taken  as  8^  lb. 


200  VOCATIONAL  MATHEMATICS 

At  atmospheric  pressure  one  cubic  foot  of  steam  has  nearly 
the  weight  of  one  cubic  inch  of  water,  and  the  weight  increases 
very  nearly  as  the  pressure.  Hence,  to  find  the  number  of  cubic 
inches  of  water  required  to  make  a  certain  amount  of  steam, 
multiply  the  number  of  cubic  feet  of  steam  by  the  absolute 
pressure  in  the  atmosphere ;  the  product  is  the  number  of 
cubic  inches  of  water  required.  In  all  such  calculations  for 
practical  purposes,  a  liberal  allowance  must  be  made  for  loss 
and  leakage. 

Absolute  pressure  is  the  total  pressure,  or  the  gauge  pressure 
plus  the  atmospheric  pressure  (which  at  sea  level  is  14.7  lb. 
per  sq.  in.). 

EXAMPLES 

1.  How  much  water  will  it  take  to  make  800  cu.  ft.  of  steam 
at  10  lb.  pressure  ? 

2.  How  much  water  will  it  take  to  make  3020  cu.  ft.  of 
steam  at  65  lb.  pressure  ? 

3.  How  much  water  will  it  take  to  make  4812  cu.  ft.  of 
steam  at  8  lb.  pressure  ? 

4.  How  much  water  will  it  take  to  make  512  cu.  ft.  of 
steam  at  75  lb.  pressure  ? 

5.  How  much  water  will  it  take  to  make  1213  cu.  ft.  of 
steam  at  80  lb.  pressure  ? 

Solve  the  following  problems  according  to  the  table  on  the 
next  page : 

6.  What  is  the  total  steam  pressure  if  the  steam  gauge  reads 
55  lb.  ? 

7.  How  many  cubic  feet  of  steam  from  2  lb.  of  water  at  a 
steam  gauge  pressure  of  65  lb.  ? 

8.  ,What  is  the  latent  heat  of  1  lb.  of  water  at  a  total 
pressure  of  75  lb.  at  307.5°  F.  ? 

9.  What  is  the  total  heat  required  to  generate  1  lb.  of  steam 
from  water  at  32°  F.  under  total  pressure  of  90  lb.  ? 


HEAT 


201 


Properties  of  Saturated  Steam 


Pbbmdeb 

VOLUMB 

Total  Heat 

Tempera- 
ture In 

Fahrenbeit 
Dejfrees 

Latent  Heat 
in  B.  T.  U. 

required  to 

XT 

ToUl 

Compared 
with  Water 

(^Hblc  Ft.  of 

Steam  fi*»ttn 

1  Lb.  of 

peneralc  1  Lb. 

of  Steum  from 

Water  at  «tio 

under  Conhtant 

Water 

I're.smire 

Heat  unltH 

0 

16 

212.0 

nm 

26.36 

966.2 

1146.1 

5 

20 

228.0 

1220 

10.72 

952.8 

1150.9 

10 

25 

240.1 

9JK5 

15.99 

945.3 

1154.6 

16 

30 

260.4 

838 

18.40 

937.9 

1157.8 

20 

36 

269.3 

726 

11.65 

931.6 

1160.5 

26 

40 

267.3 

640 

10.27 

926.0 

1162.9 

30 

46 

274.4 

672 

9.18 

920.9 

1165.1 

86 

60 

281.0 

618 

8.31 

910.3 

1167.1 

40 

66 

287.1 

474 

7.61 

912.0 

1169.0 

46 

60 

292.7 

437 

7.01 

908.0 

1170.7 

60 

65 

298.0 

405 

6.49 

904.2 

1172.3 

66 

70 

302.9 

378 

6.07 

900.8 

1173.8 

60 

76 

307.5 

353 

5.68 

897.5 

1175.2 

66 

80 

312.0 

333 

5.35 

894.3 

1176.6 

70 

86 

316.1 

314 

5.05 

891.4 

1177.9 

76 

90 

320.2 

298 

4.79 

888.5 

1179.1 

80 

95 

324.1 

283 

4.55 

885.8 

1180.3 

86 

100 

327.9 

270 

4.33 

888.1 

1181.4 

00 

106 

331.3 

267 

4.14 

880.7 

1182.4 

96 

110 

334.6 

247 

3.97 

878.3 

1183.6 

100 

116 

338.0 

237 

3.80 

875.9 

1184.5 

110 

125 

344.2 

219 

3.50 

871.5 

1186.4 

120 

135 

350.1 

203 

3.27 

867.4 

1188.2 

130 

146 

355.6 

liK) 

3.06 

863.5 

1189.9 

140 

165 

361.0 

179 

2.87 

859.7 

1191.6 

160 

166 

366.0 

169 

2.71 

866.2 

1192.9 

Steam  Heating 

A  steam  heating  system  with  steam  having  a  pressure  of  less 
than  15  lb.  by  the  gauge  is  called  a  low  pressure  system.  If 
the  steam  pressure  is  higher  than  15  lb.  it  is  called  a  high 
pressure  system.  When  the  water  of  condensation  flows  back 
to  the  boiler  by  gravity  alone,  the  apparatus  is  known  as  a 
gravity  circulating  system.  When  the  boiler  is  run  at  a  high 
pressure  and  the  heating  system  at  a  low  pressure,  the  con- 
densed steam  must  be  returned  to  the  boiler  by  a  pump,. steam 
return  trap,  or  injector. 


202  VOCATIONAL  MATHEMATICS 

The  quantity  of  heat  given  off  by  the  radiators  of  steam  pipes,  in  the 
ordinary  methods  of  heating  buildings  by  direct  radiation,  varies  from  If 
to  3  heat  units  per  hour  per  square  foot  of  radiating  surface  for  each  degree 
of  difference  in  temperature ;  an  average  of  from  2  to  2i  is  a  fair  estimate. 

One  pound  of  steam  at  about  atmosplieric  pressure  contains  1146  heat 
units,  and  if  the  temperature  in  the  room  is  to  be  kept  at  70"^  F.,  while  the 
temperature  of  the  pipes  is  212  degrees,  the  difference  in  temperature  is 
142  degrees.  Multiplying  this  by  2J,  the  emission  of  heat  will  be  319^  heat 
units  per  hour  per  square  foot  of  radiating  surface.  A  rule  often  given  is 
to  allow  one  square  foot  of  heating  surface  in  the  boiler  for  every  eight  to 
ten  square  feet  of  radiating  surface. 

In  steam  heating  the  following  rule  is  used:  To  find  the 
amount  of  direct  radiating  surface  required  to  heat  a  room, 
basing  the  calculation  upon  its  cubic  contents,  allow  one  square 
foot  of  direct  radiating  surface  for  the  cubic  feet  shown  in 
the  following  table. 

Proportion  of  Radiating  Surface  to  Volume  of  Rooms 

Bathrooms  or  living  rooms,  with  2  or  3  exposures   .     .     .  40  cu.  ft. 

Living  rooms,  with  1  or  2  exposures 50  cu.  ft. 

Sleeping  rooms        65-70  cu.  ft. 

Halls 50-70  cu.  ft. 

Schoolrooms 60-80  cu.  ft. 

Large  churches  and  auditoriums 65-100  cu.  ft. 

Lofts,  workshops,  and  factories 75-150  cu.  ft. 

The  above  ratios  will  give  reasonably  good  results  on  ordi- 
nary work  if  proper  judgment  is  exercised. 

EXAMPLES 

1.  How  much  radiating  surface  is  necessary  to  heat  a  bath- 
room containing  485  cu.  ft.  ? 

2.  How  much  radiating  surface  is  necessary  to  heat  a  bath- 
room 10^'  X  5i'  X  9'  ? 

3.  How  much  radiating  surface  is  necessary  to  heat  a  living 
room  with  three  windows,  and  containing  2798  cu.  ft.  ? 

4.  How  much  radiating  surface  is  necessary  to  heat  a  living 
room  18'  X  16|-'  X  10',  with  three  windows? 


CHAPTKK   XIII 
BOILERS 

There  are  two  classes  of  boilers  —  water  tube  and  fire  tube. 
The  difference  between  the  two  is  that  water  flows  through 
water  tube  boilers  and  the  fire  to  heat  the  water  is  outside, 
while  in  the  fire  tube  boiler  the  conditions  are  reversed. 

Return  Tubular  Boilers.  —  The  boiler  most  widely  used  in 
America  is  the  return  tubular,  which  is  a  type  of  fire  tube 
boiler.  It  is  a  closed  tube,  simple  in  construction,  inexpensive 
to  make,  and  easy  to  clean  and  repair.  The  first  horizontal 
tubular  boiler  was  an  ordinary  storage  tank  made  of  iron. 
Now  horizontal  tubular  boilers,  16  to  20  feet  long,  and  4  to  8 
feet  in  diameter,  and  even  larger,  are  used  and  can  withstand 
a  pressure  of  150  lb.  per  sq.  in. 

Boilers  up  to  fourteen  feet  in  length  are  constructed  of  two  plates,  each 
forming  the  entire  circumference  of  the  boiler.  Above  fourteen  feet  long 
the  shell  is  constructed  of  three  plates  which  make  the  required  length  of 
the  boiler  shell.  These  plates  are  J,  f,  ^,  or  ^  in.  thick,  having  from 
45,000  to  86,000  lb.  tensile  strength. 

Tensile  Strength.  —  The  tensile  strength  is  the  pull  applied 
in  the  direction  of  its  length  required  to  break  a  bar  of 
boiler  plate  one  square  inch  in  area.  Different  pieces  are 
taken  from  the  various  parts  of  the  boiler  plate,  reduced  to  \ 
inch  square,  and  subjected  to  pressure  on  a  testing  machina 
The  average  strength  of  the  samples  is  thus  obtained  and  multi- 
plied by  16  to  determine  the  strength  of  one  square  inch.  The 
tensile  strength  is  usually  stamped  on  the  boiler  steel.  If  it  is 
not  stami)ed  on  it,  the  tensile  strength  is  48,000  lb. 

203 


§ 

as  45 


03     g 

^   a 

mi 

en 
ts    as 


pd    c3 

O 

N 

o 
W 


BOILERS 


205 


If  four  samples  give  tests  of  3998  lb.,  4001  lb.,  4001  lb.,  and  4000  lb., 
then  the  avera«;e  is  4000.  Therefore,  for  one  square  inch  the  tensile 
strength  is  4000  x  10  (the  numlu'r  of  quartt-r  sq.  in.  in  one  aq.  in.)  = 
64,000  lb. 


O 


^######JJ    i 


SlNOLK    RiVETKD   LAP  JoiVT  ;  ^   ErFirTEXCY  ABOUT   56% 


o 


#--###  #H 


Double  Riveted  Lap  Joint;  Efficiency  about  70% 


Triple  Riveted  Butt-strap  Joint;  Efficiency  about  85% 

Safe  Working  Pressure.  —  In  order  to  know  the  safe  working 
pressure  of  a  single  riveted  boiler,  it  is  necessary  to  multiply 
one  sixth  of  the  tensile  strength  by  the  thickness  of  the  shell, 
and  divide  this  product  by  the  inside  radius  of  the  shell.     If 


1  The  efficiency  of  a  riveted  joint  id' the  ratio  of  the  strength  of  the  joint  to 
that  of  the  solid  plate. 


206  VOCATIONAL   MATHEMATICS 

a  boiler  is  double  riveted,  add  20  per  cent  to  the  safe  pressure 
of  a  single  riveted  boiler  of  the  same  dimensions. 

Example.  —  What  is  the  safe  working  pressure  of  a  single 
riveted  boiler  having  72"  diameter,  i"  shell,  if  the  boiler  plate 
has  a  tensile  strength  of  66,000  lb.  ? 

I  X  66,000=  11,000 
11,000  X  .5"  =  5500 
5500  H-  36"  (radius)  =  152|  lb.  approx.    Ans. 

EXAMPLES 

1.  (a)  What  is  the  safe  working  pressure  of  a  single  riveted 
boiler  of  60"  diameter,  j%''  shell,  the  T.  S.  being  60,000  lb. 
per  sq.  in.  ?  (b)  What  is  the  safe  working  pressure  of  a  double 
riveted  boiler  of  the  same  dimensions? 

2.  (a)  What  is  the  safe  working  pressure  of  a  single  riveted 
boiler  of  54''  diameter,  f "  shell,  the  T.  S.  being  60,000  lb.  ? 
(6)  What  is  the  safe  working  pressure  of  a  double  riveted 
boiler  of  the  same  dimensions  ? 

3.  (a)  What  is  the  safe  working  pressure  of  a  single  riveted 
boiler  of  48"  diameter,  f "  shell,  the  T.  S.  being  60,000  lb.  ? 
(b)  What  is  the  safe  working  pressure  of  a  double  riveted 
boiler  of  the  same  dimensions? 

4.  (a)  What  is  the  safe  working  pressure  of  a  single  riveted 
boiler  of  42"  diameter,  j\"  shell,  the  T.  S.  being  60,000  lb.  ? 
(6)  Of  a  double  riveted  boiler  of  the  above  dimensions? 

The  Plate. — The  diameter  of  the  rivet  used  for  boiler  plate 
is  generally  double  the  thickness  of  the  plate.  The  distance 
between  the  rivet  holes  of  a  boiler  is  found  by  dividing  the 
area  of  the  rivet  hole  by  the  thickness  of  the  plate.  The  pitch 
or  distance  between  the  rivet  hole  centers  in  a  boiler  plate 
is  found  by  dividing  the  area  of  the  rivet  hole  by  the  thick- 
ness of  the  plate  and  adding  the  diameter  of  one  hole.  The 
pitch  in  a  rivet  hole  is  found,  when  the  shearing  strength  is 


208  VOCATIONAL  MATHEMATICS 

known,  by  multiplying  the  area  of  the  rivet  hole  by  the  shear- 
ing strength,  then  multiplying  the  thickness  of  the  plate  by 
the  tensile  strength,  dividing  the  first  product  by  this  product, 
and  adding  one  rivet  hole  diameter  to  the  quotient. 

Example.  —  What  thickness  of  plate  should  be  used  on  a 
40-inch  boiler  to  carry  125  lb.  pressure,  if  the  tensile  strength 
of  the  plate  is  60,000  lb.  ? 

125  =  steam  pressure 
6  =  factor  of  safety 
40  =  diameter  of  boiler 
20  =  i  diameter  of  boiler 
60,000  =  tensile  strength  of  plate 
126  X  6  =  750 
750  X  20  =  15,000 

— 5 —  =  .25  or  i"  thickness  of  plate.    Ans. 
60,000  *  ^ 

The  Boiler  Inspection  Department  of  Massachusetts  recom- 
mends the  following  fornuila  for  determining  the  thickness  of 
boiler  plate : 

rp^PxRx  F.S. 
T.S,  X  % 

T  =  thickness  of  plate 

P  =  pressure 

E  =  radius  (|  diameter  of  boiler) 
F.  S.  -  safety  factor 
T.  S.  =  tensile  strength 

%  =  strength  of  joint 

Example. — What  thickness  of  plate  should  be  used  on  a 
40-inch  boiler  to  carry  125  lb.  pressure,  if  the  strength  of  the 
plate  is  60,000  lb.,  using  50  %  as  the  strength  of  the  joint? 

r  =  ^f><^Qxe  =  |// sheet.     Ans. 
60,000  X  .50       ^ 

Find  the  safe  working  of  the  same  boiler  with  the  above 

figures : 

Safe  working  pressure  =  60,000  x  .5  x  .50  ^  ^35  lb.     Ans. 

20  X  6 


BOILERS 


209 


Small  Vertical  Boilkr 


Size.  —  The   engineer   often   has  to  calculate  the  size  of  a 
boiler  to  carry  a  definite  steam  pressure. 

The  size  of  a  single  riveted  l)oiler  may  be  found  by  multi- 


210  VOCATIONAL  MATHEMATICS 

plying  J  of  the  tensile  strength  by  the  thickness  of  the  shell, 
and  dividing  this  product  by  the  steam  pressure.  The  quo- 
tient is  the  radius  of  the  boiler.  Multiply  this  radius  by  two 
to  obtain  the  diameter  (twice  this  radius  equals  the  theoretical 
diameter)  ;  add  one  fifth  of  the  diameter  just  found  as  a  safety 
factor,  and  this  sum  gives  the  working  diameter  of  a  boiler 
that  will  safely  carry  the  required  pressure. 

Example.  —  What  is  the  diameter  of  a  ^Ijoiler  that  will  with- 
stand 150  lb.  pressure,  if  made  of  f"  steel,  60,0U0  lb.  T.  S.? 

I  X  60,000  =  10,000 
10,000- X  f  =  3750 

-^■^-^Jt  =  2.j"  tlieoretical  radius 
50'  =  theoretical  diameter 
50"  4-  10"  =  60"  working  diameter.    A7is. 

The  Boiler  Inspection  Department  of  Massachusetts  recom- 
mends the  following  formula  for  finding  the  diameter  of  a  boiler 
when  the  pressure,  thickness,  tensile  strength,  and  per  cent  are 
known. 

rf^Tx  T.S.x  %      2 
P.F. 

D  =  diameter  of  boiler 
T  =  thickness  of  plate 
T.  S.  =  tensile  strength 
0/,  =  strength  of  joint 
P  =  pressure 
F  =  safety  factor 

Example.  — What  is  the  diameter  of  a  boiler  having  |"  shell, 
allowing  50  per  cent  for  the  strength  of  the  joint,  with  a  tensile 
strength  of  60,000  lb.,  when  the  factor  of  safety  is  6  and  the 
pressure  of  steam  is  125  lb.  ? 

^  .l><JiO,OjOO  X  ,50  ^  2  ^  4^„  diameter.     Ans. 
125  X  6 


B01LKR8  211 

EXAMPLES 

1.  What  is  the  diameter  of  the  rivet  for  a  plate  J"  thick  ? 

2.  How  much  space  is  there  between  the  rivet  hole^,  \^"  in 
diameter,  on  a  J^"  plate  ? 

3.  What  is  the  pitch  in  a  \"  boiler  plate,  if  the  rivet  diam- 
eter is  f"  and  the  hole  diameter  is  |J"? 

4.  What  is  the  working  diameter  of  a  boiler  made  of  |" 
steel,  60,000  lb.  T.  S.,  that  will  withstand  90  lb.  pressure? 

5.  What  thickness  of  plate  should  be  used  on  a  72"  boiler 
with  a  T.  S.  of  66,000  lb.  to  carry  a  pressure  of  90  lb.  ? 

6.  How  much  space  is  there  between  rivet  holes  J"  in 
diameter  on  a  ^^"  plate  ? 

7.  W^hat  is  the  working  diameter  of  a  boiler  made  of  j\" 
steel,  60,000  lb.  T.  S.,  that  will  withstand  150  lb.  pressure? 

8.  What  is  the  pitch  of  a  rivet  hole  of  a  f "  boiler  plate 
with  a  diameter  of  ||",  a  shearing  strength  of  80,000  lb., 
and  T.  S.  of  60,000  lb.  ? 

9.  What  is  the  working  diameter  of  a  boiler  made  of  J" 
steel,  60,000  lb.  T.  S.,  that  will  withstand  125  lb.  pressure  ? 

10.  What  should  be  the  diameter  of  a  rivet  to  be  used  in  a 
A"  plate? 

11.  W^hat  is  the  pitch  in  a  V'  lx)iler  i)late  with  a  rivet  diam- 
eter of  I"  and  a  rivet  hole  diameter  of  |J"? 

12.  What  thickness  of  plate  should  be  used  on  a  48"  boiler 
to  carry  60  lb.  pressure,  with  a  T.  S.  of  60,000  lb.  ? 

13.  What  is  the  working  diameter  of  a  boiler  made  of  |" 
steel,  60,000  lb.  T.  S.,  that  will  withstand  110  lb.  pressure  ? 

14.  What  is  the  pitch  in  a  rivet  hole  |"  in  diameter  in  a 
boiler  with  a  shearing  strength  of  32,000  lb.,  if  the  plate  has 
a  tensile  strength  of  60,000  lb.,  and  is  ^j^"  in  thickness  ? 

15.  What  is  the  working  diameter  of  a  boiler  made  of  ,^5" 
steel,  60,000  lb.  T.  8.,  that  will  withstand  60  lb.  pressure? 


212  VOCATIONAL  MATHEMATICS 

Boiler  Tubes 

A  boiler  tube  is  open  at  both  ends.  Tiierefore  it  is  not 
necessary  to  consider  the  area  of  the  bases  in  computing  the 
heating  surface  of  a  tube.  The  area  of  the  cylindrical  surface 
is  all  that  is  necessary  to  find.  Boiler  tubes  are  often  meas- 
ured in  terms  of  the  heating  surface  per  foot. 

Example.  —  What  is  the  heating  surface  per  foot  for  a  3" 
tube,  1"  thick? 

3"  =  diam  -of  tube 
2  X  ^"  =  \"  OT  twice  thickness  of  tube 
S-\"  =  2f"  inside  diam.  of  tube  =  2.75" 
2.75"  X  3.1416  =  8.6394"  inside  circumference 
8.6394"  X  12"  =  103.6728  sq.  in. 
103.6728 


144 


=  .7199  sq.  ft.  heating  surface 


To  find  the  total  tube  heating  surface  of  a  boiler : 
Multiply  the  heating  surface  per  foot  of  length  by  the  length 
of  the  tubes  in  feet  and  that  product  by  the  number  of  tubes 
in  the  boiler. 

Example.  —  If  a  boiler  which  has  110  3"  tubes  y  thick  is 
12'  between  front  and  back  heads,  what  is  its  tube  heating 
surface  ? 

.72  =  heating  surface  per  foot  length 
12  =  length  of  tubes  in  feet 
110  =  number  of  tubes 
.72  X  12  =  8.64  sq.  ft.  per  tube 
8.64  X  110  =  950.4  sq.  ft.  tube  heating  surface.    Ans. 

Example.  —  In  the  above  example  what  percentage  of  the 
total  volume  of  the  boiler  do  the  tubes  represent  if  the  di- 
ameter is  60"  ? 

60  =  diam.  of  shell 
60  X  60  =  3600 
3600  X  .7854  =  2827.44  sq.  in.  area 

^^^1^  =  19.635  sq.  ft.  area 
144 

19.64  X  12'  =  235.68  cu.  ft.  volume  of  shell 


BOILERS  213 

8"  =  diam.  of  tube  3x8  =  9 
9  X  .7864  =  7.0086  sq.  in.  area 

"^-^^^  =  .049  sq.  ft.  area 
144  ^ 

.049  X  1  =  cu.  ft.  for  1  ft.  of  tube  length 

.049  X  12'  =  0.588  cu.  ft.  (capacity  of  each  tube) 

.688  X  110  =  04.08  cu.  ft.  occupied  by  tubes 

64J  tubes  volume  ^  ^7  =  27  per  cent  shell  volume.   Arts. 
235.7  shell  volume 

Fusible  Plug.  —  A  fusible  plug  is  a  bi-ass  plug  with  a  tapering  center  of 
Banca  tin.  The  Lirge  end  is  put  next  to  the  pressure  to  prevent  the  soft 
metal  from  blowing  out.  This  plug  is  screwed  into  the  rear  head  of  a 
boiler  not  less  than  two  inches  above  the  top  row  of  tubes  and  should 
extend  one  inch  into  the  water  to  prevent  it  from  becoming  scaled.  If 
the  water  falls  below  this  plug  the  soft  metal  melts,  allowing  the  steam  to 
escape,  thus  giving  warning. 

Manhole.  —  A  manhole,  oval  in  shape,  is  put  in  the  top  or  in  the  heads 
of  the  boiler,  to  allow  a  person  to  enter  the  inside  to  inspect  the  boiler. 
It  is  made  tight  by  the  use  of  a  rubber  gasket. 

Hand  Hole  and  Blow-off.  —  Hand-hole  plugs  are  put  into  the  bottom  of 
the  front  and  rear  heads  of  a  boiler  to  permit  washing  out.  The  blow-off 
is  connected  at  the  bottom  of  the  shell  at  the  rear  end  with  a  valve  on  the 
pipe  outside  the  brickwork,  called  a  blow-off  valve.  This  is  to  empty  the 
boiler,  or  to  blow  it  down  one  gauge.  It  is  necessary  to  blow  down 
the  boiler  each  morning  in  order  to  rid  it  of  the  .sediment  that  has  settled 
at  the  bottom  each  day.  Boilers  should  be  entirely  emptied  and  washed 
out  at  least  once  a  month ;  the  necessity  for  this  is  determined  by  the 
quality  of  the  feed  water. 

Water  Gauge.  — The  water  gauge  registers  the  height  of  the  water  in 
a  boiler.  It  consists  of  a  small  cast  iron  drum  placed  in  an  upright  posi- 
tion in  front  of  the  boiler  and  provided  with  a  gla.ss  gauge,  cocks,  water, 
and  steam  connections.  The  pipe  connections  are  arranged  so  that  dry 
steam  enters  the  top  and  water  the  bottom,  with  a  blow-off  valve  for  the 
water  column  and  gauges.  The  water  should  be  kept  up  to  the  second 
gauge  while  the  boiler  is  working,  and  up  to  the  third  gauge  at  night. 
The  first  duty  of  the  engineer  is  to  see  that  the  water  is  at  the  proper 
level.  To  be  sure  that  the  glass  is  registering  correctly,  the  gauge  cocks 
have  to  be  tried.  One  of  the.se  is  below  the  water  line  and  one  above  it. 
If  the  water  in  the  boiler  is  right,  steam  will  come  out  of  the  upper  one 
and  water  out  of  the  lower ;  if  the  water  in  the  boiler  is  too  low,  steam 
will  come  out  of  both. 


214 


VOCATIONAL  MATHEMATICS 


Safety  Valves 

The  power  of  a  boiler  depends  upon  the  amount  of  heating 
surface  that  the  boiler  contains.  As  the  cylinder  of  the  boiler 
is  made  to  withstand  a  certain  pressure,  an  excess  may  cause 
it  to  explode.  So  it  is 
necessary  that  the  engi- 
neer should  know  when 
the  pressure  is  exceeded. 
Various  devices  to  b* 
attached  to  boilers  have 
been  invented  to  give 
warning.  One  of  these  is 
the  safety  valve. 

Every  boiler  should  have  at 
least  two  safety  valves,  a  water 
gauge,  and  a  pressure  gauge. 
The  function  of  a  safety  valve 
is  to  relieve  the  boiler  of  all 
pressure  in  excess  of  that  at 
which  the  valve  is  set  to  blow 
off.  It  is  placed  at  the  top  of 
the  boiler  and  piped  outside.  The  careful  engineer  tries  the  safety  valve 
every  day  to  see  if  it  is  in  working  order. 

The  size  of  the  safety  valve  is  very  important.  The  area  of 
the  grate,  the  weight  of  the  fuel  burned,  and  the  steam  pres- 
sure have  to  be  considered  when  calculating  the  size  of  the 
valve.  The  amount  of  steam  generated  in  a  given  time  and 
the  pressure  caused  by  the  steam  will  depend  upon  the  weight 
of  coal  burned.  The  velocity  of  the  escape  of  the  steam 
through  the  valve  will  depend  upon  the  pressure  of  the  steam. 
A  low  pressure  safety  valve  is  not  higher  than  thirty  pounds. 
The  figure  stamped  on  the  lever  of  the  safety  valve  shows  the 
limit  of  pressure. 

Lever  Safety  Valve,  —  The  lever  safety  valve  is  placed  on  an 
opening  in  the  top  of  the  boiler.     This  valve  consists  of  a  disk, 


Steam  Pressure  Gauge 


BOILERS 


215 


Sufety  Valve 


a  stem,  and  a  lever  whicli  has  a  weight  hung  on  the  end.  The 
weight  keeps  the  valve  in  place  until  the  pressure  of  the  steam 
in  the  boiler  overcomes  that  of  the  weight  so  that  the  valve  is 
jmshed  up  and  gives  warning  that  the  jiressure  of  the  steam 
must  be  lessened  to  prevent  an 
explosion. 

C  =  fulcrum  CzZ3 

A  =  distance  from  fulcrum  to  center 

of  valve 
B  =  distance  from  fulcmm  to  center 

of  weight 
P=  pressure  (total) 
ir=  weight  BW  =  AP 

Bx  ir-^  P  =  A 
Ax  P^]V=B 
Px  A-^  B  =  ]V 
Wx  B^  A  =  P 

Pop  Safety  Valve.  —  The  pop  safety  valve  has  displaced  the 
ever  safety  valve  to  a  large  extent. 

The   size  of  the  safety  valve  that   should  be  placed  on  a 
boiler  may  be  determined  by  the  following  rule : 

Grate  surface  x  22.5  (1) 

Gauge  pressure  +  8.62  (2) 

(1)  -i-  (2)  =  area  of  valve 
.  ^  22.5  G 
P+8.62 
Divide  area  of  valve  by  .7854  and  extract  the  square  root  of 
the  quotient.     The  result  is  the  diameter. 

Example. — What  shall  be  the  diameter  of  a  safety  valve  on  a 
boiler  with  36  sq.  ft.  of  grate  surface  carrying  100  lb.  pressure  ? 

36  X  22.5  =  810 
100  +  8.62  =  108.62 


810 
108.62 

\.7864 


=  7.6  sq.  in.,  area  of  valve 
=  3 J"  approx.    Ana. 


216  VOCATIONAL  MATHEMATICS 

The  Boiler  Inspection  Department  of  Massachusetts  recom- 
mends the  following  formula  for  calculating  the  size  of 
safety  valve. 

Diameter  of  valve  =  yj^^J^  x  11-^.7854. 

D  =  diameter  of  valve 

W  =  weight  of  water  in  pounds  evaporated  per  square  foot  of  grate 

surface  per  second 
P  =  pressure  absolute  at  which  safety  valve  is  set  to  blow 


EXAMPLES 

1.  What  must  be  the  size  of  a  safety  valve  on  a  boiler  with 
48  sq.  ft.  of  grate  surface,  carrying  98  lb.  pressure  ? 

2.  The  diameter  of  a  boiler  tube  is  4"  and  the  length  18'. 
Find  the  area  of  the  surface  of  the  tube  in  sq.  ft. 

3.  What  must  be  the  size  of  a  safety  valve  on  a  boiler  with 
42  sq.  ft.  of  grate' surf  ace,  carrying  84  lb.  pressure? 

4.  The  diameter  of  a  boiler  tube  is  3^"  and  its  length  16'. 
What  is  its  area  ? 

5.  The  diameter  of  a  lever  safety  valve  of  a  steam  boiler 
is  3",  the  length  of  the  lever  from  its  fulcrum  to  a  jjoint  at 
which  a  50  lb.  weight  is  suspended  is  24",  and  the  distance 
from  the  fulcrum  to  the  point  where  the  lever  in  a  horizontal 
position  presses  upon  the  valve  is  3".  At  what  steam  pressure 
per  sq.  in.  will  the  boiler  blow  off  ? 

6.  What  is  the  total  tube  heating  surface  of  a  boiler  having 
112  3"  tubes,  each  ^"  thick,  if  the  distance  between  the  front 
and  back  heads  is  16'  ? 

7.  On  a  certain  boiler  the  safety  valve  lever  is  24"  long 
and  weighs  3  lb.  and  carries  at  its  extremity  a  weight  of  30  lb. 
The  length  from  the  fulcrum  of  the  lever  to  the  valve  spindle 
is  3",  and  from  the  fulcrum  to  the  center  of  gravity  of  the 
lever  16".     If  the  valve  has  an  area  of  10  sq.  in.  and  weighs 


BOILERS  217 

with  its  spindle  1^  lb.,  at  what  steam  pressure  per  sq.  in.  will 
the  boiler  blow  off  ? 

8.  In  Example  7  if  it  is  desired  to  reduce  the  maximum 
steam  pressure  by  one  half,  at  what  point  on  the  lever  must 
the  weight  be  hung  ? 

9.  What  is  the  heating  surface  per  foot  of  a  2^"  boiler 
tube  ^"  thick  ? 

10.  What  is  the  total  tube  heating  surface  of  a  boiler  having 
90  3f  tubes,  each  ^  thick,  if  the  boiler  is  17'  long? 

11.  What  is  the  total  tube  heating  surface  of  a  boiler  having 
80  3"  tubes,  each  |"  thick  and  16'  long  ? 

12.  What  is  the  percentage  of  the  total  tube  volume  to  the 
total  boiler  volume  of  a  boiler  72"  in  diameter,  with  shell  18' 
long,  having  70  tubes,  each  4"  in  diameter  ? 

13.  What  is  the  total  tube  heating  surface  of  a  16'  boiler 
having  60  3"  tubes,  each  ^"  thick  ? 

14.  WTiat  is  the  heating  surface  per  foot  of  a  2^'  boiler  tube 
V  thick? 

15.  The  safety  valve  of  a  boiler  is  4"  in  diameter,  the 
center  of  the  valve  is  5"  from  the  pin  at  the  end  of  the 
lever,  the  lever  is  51"  long  from  the  pin  and  carries  a 
weight  of  112  lb.  at  the  end  ;  the  weight  of  the  valve  is  7}j  lb., 
of  the  lever  42  lb. ;  the  center  of  gravity  of  the  lever  is  16" 
from  the  pin.     At  what  pressure  will  the  valve  blow  off  ? 

Superheated  Steam 

The  steam  used  in  boilers  should  be  as  dry  as  possible.  It 
may  be  made  dry  by  heating  it  to  a  higher  temperature  by 
passing  it  through  a  vessel  or  coils  of  pipe  separated  from  the 
boiler  and  called  a  superheater.  Every  passage  conveying 
superheated  steam  must  be  well  covered  with  non-conducting 
material. 


218  VOCATIONAL  MATHEMATICS 

The  temperature  of  steam  in  contact  with  the  water  from 
which  it  is  generated,  as  in  the  ordinary  steam  boiler,  depends 
upon  the  pressure.  But  if  the  vessel  is  closed,  as  in  the  case 
of  boilers,  the  pressure  becomes  greater  and  raises  the  boiling 
point  of  the  water.  Steam  confined  and  under  pressure  has 
considerable  energy  stored  up  and  is  a  powerful  moving  force 
when  allowed  to  enter  the  piston  chamber  of  an  engine. 

Boiler  Pumps 

The  water  inside  a  boiler  is  usually  kept  at  the  proper  level 
by  means  of  pumps  or  injectors.  Most  boilers  have  at  least 
two  means  of  feeding  water.  Steam  pumps  are  most  com- 
monly used  on  stationary  and  marine  boilers.  There  are  sev- 
eral kinds  of  steam  pumps:  such  as  boiler  feeders,  general 
surface  pumps,  tank  pumps,  and 
municipal  waterworks  pumps. 
These  are  single  or  duplex. 

The  duplex  pump  is  most  com- 
monly used  because  it  is  the 
simplest.  The  mechanism  of  it 
is  much  the  same  as  that  of  the 
ordinary  force  pump,  although  it 
is  a  combination  of  two  steam 
pumps,  so  arranged  that  the 
valve  of  one  is  operated  by  the 
piston  of  the  other. 

An  injector  is  an  apparatus  for 

forcing   water    against    pressure 

by  the  direct  action  of  steam  on 

,r  ,  Ti.  •  •  n  J       Sectional  View  of  Injector 

the  water,    it  is  universally  used 

on  locomotives  and  occasionally  on  stationary  boilers.  Steam 
is  led  from  the  boiler  through  a  pipe  which  terminates  in  a 
nozzle  surrounded  by  a  cone.  This  cone-shaped  pipe  is  con- 
nected with  the  water  tank  or  well  where  the  water  is  stored. 


BOILERS  219 

When  the  steam  passes  into  the  injector,  it  rushes  with  great 
velocity  from  the  nozzle  and  creates  a  partial  vacuum  in  the 
cone.  This  causes  atmospheric  pressure  to  force  water  up  to 
the  cone,  and  there  the  kinetic  action  of  the  steam  imparts 
velocity  to  it  and  overcomes  the  boiler  pressure. 

Size  of  Pump.  —  To  find  the  size  of  a  pump  to  supply  a  boiler : 
Multiply  the  volume  in  cu.  in.  of  the  water  evaporated  by  each 
H.  P.  per  hour  by  the  number  of  H.  P.,  and  divide  this  product 
by  the  numl)er  of  inches  the  plunger  travels  per  minute.  The 
quotient  gives  the  area  or  size  of  the  pump. 

Example.  —  Find  the  size  of  a  pump  that  supplies  a  boiler 
furnishing  steam  for  a  50  H.  P.  engine,  the  plunger  making 
47 j^  ft.  per  minute,  if  30  lb.  of  water  are  evaporated  per  hour 
for  each  H.  P. 

30  lb.  water  =  .'U  gal.  approx. 
H\  X  231  =  808^  cu.  in. 
47i  X  12  =  570 
60  X  808.5  -r-  570  =  70.J)2  sti.  in.  in  area.     Ans. 

The  H.  P.  necessary  to  pump  a  given  amount  of  water  per 
minute  to  a  given  height  is  found  by  multiplying  the  total 
weight  of  the  water  in  pounds  by  the  height  in  feet  and  divid- 
ing by  33,000 ;  the  quotient  will  be  the  required  II.  P. 

Capacity  of  Pump.  —  To  find  the  number  of  gallons  of  water 
that  a  pump  of  a  given  size  is  capable  of  raising  per  minute, 
nmltiply  the  area  of  the  water  cylinder  in  square  inches  by 
the  number  of  inches  the  piston  travels  per  minute,  divide  the 
result  by  231,  and  the  quotient  will  be  the  number  of  gallons 
per  minute. 

Example.  —  If  the  water  end  of  a  pump  has  a  diameter  of 
6"  and  the  pump  is  running  at  a  speed  of  100'  per  minute, 
if  no  leaks  are  accounted,  how  many  gallons  of  water  is  it 
delivering  ? 


220  VOCATIONAL    MATHEMATICS 

Area  of  pump  =  6""  x  .7864  =  28.27  sq.  in. 
100  ft.  or  1200  in.  per  minute 
28.27  X  1200  =  33,924  cu.  in. 

^M^  -  146.8  gal.  per  minute.    Ans. 
231  ^ 

Example.  —  How  much  water  Avill  a  pump  of  4  in.  diameter 

deliver  in  one  hour  if  the  plunger  makes  47^  ft.  per  minute  ? 

4  X  4  X  .7854  =  12.56  sq.  in. 
12.56  X  47.5  X  12  =  7159.2  cu.  in. 
7159.2  X  60 


231 


1859.5  gallons  per  hour 


EXAMPLES 

1.  Find  the  size  of  a  pump  supplying  a  boiler  which  fur- 
nishes steam  for  80  H.  P.  in  1  hour's  time,  if  the  plunger 
makes  55  ft.  per  minute  and  30  lb.  of  water  are  evaporated 
per  hour  for  each  H.  P. 

2.  How  much  water  will  a  pump  deliver  in  one  hour,  if 
the  size  is  S"  drain  and  the  plunger  makes  43'  per  minute  ? 

3.  How  many  gallons  of  water  will  a  6''  pump  with  a  piston 
speed  of  60'  per  minute  raise  in  one  hour  ? 

4.  What  H  P.  is  necessary  to  pump  7583  gallons  of  water 
134  feet  in  19  minutes  ? 

5.  Find  the  number  of  gallons  of  water  a  pump  with  an 
end  diameter  of  4''  and  running  at  the  rate  of  84'  per  minute 
will  deliver. 

6.  Find  the  size  of  a  pump  that  supplies  a  boiler  furnishing 
steam  at  98  H.  P.  in  one  hour's  time,  the  plunger  making  72  ft. 
per  minute,  if  30  lb.  of  water  are  evaporated  per  hour  for 
each  H.  P. 

7.  How  much  water  will  a  pump  deliver  in  one  hour,  if 
the  size  of  the  drain  is  3^"  and  the  plunger  makes  39|^'  per 
minute  ? 

8.  Find  the  number  of  gallons  of  water  that  a  5y  pump 
with  a  piston  speed  of  54'  per  minute  will  raise  in  one  hour. 


BOILERS  221 

9.  What  H.  P.  is  necessary  to  pump  2981  gallons  47  feet 
high  in  34  minutes  ? 

10.  How  much  water  will  a  pump  deliver  in  30  minutes  if 
the  drain  is  4}"  and  the  plunger  makes  51  J'  per  minute  ? 

11.  Find  the  number  of  gallons  of  water  a  pump  with  a 
diameter  of  3"  and  running  at  rate  of  109'  per  minute  will 
deliver. 

12.  What  H.  P.  is  necessary  to  pump  21,809  gallons  of 
water  60  feet  in  1  hour  and  13  minutes  ? 


CHAPTER   XIV 


CONNECTING 
ROD  — -" 


CROSS  HEAD- 


PISTON    ROD 


GUIDE  BARS 


ENGINES 

The  principal  parts  of  a  simple  engine  are  the  frame,  cyl- 
inder, piston,  rods,  eccentric,  crank,  shaft,  and  governor.  The 
cylinder  is  the  long,  round, 
iron  band  or  tube  in  which 
the  piston  works.  The  piston 
is  a  device  fitting  into  the 
cylinder  and  dividing  it  into 
compartments.  Packing  rings 
are  provided  to  make  it  steam 
tight.  The  piston  moves  back 
and  forth,  forced  by  the  steam 
which  is  alternately  admitted 
on  each  side  of  it  by  means  of 
valves.  This  back  and  forth 
movement  thus  imparted  to 
the  piston  by  the  steam  is 
transmitted  to  the  crank  and 
then  to  the  large  flywheel. 
The  flywheel,  by  means  of  a 
belt  or  cable,  transmits  motion 
to  smaller  wheels  or  pulleys 
which  drive  machines. 

After  the  steam  has  moved  the  piston  either  way,  it  escapes 
into  the  air  or  passes  into  one  or  more  cylinders,  where  it  re- 
ceives further  expansion.  An  engine  which  allows  steam  to 
escape  into  one  cylinder  only  is  called  a  simple  engine ;  if  the 
steam  is  allowed  to  expand  twice,  it  is  a  compound  engine; 
and  if  three  times,  a  triple  expansion  engine. 

222 


PIPE  TO  BOILER 


SLIDE  VALVE- 


STEAM  CHEST 


Vertical  Steam  Enqine 


ENGINES 


223 


Maximum  Pressure.  —  The  pressure  on  the  guides,  exerted 
by  the  crosshead  at  right  angles  to  the  line  of  the  piston  rod, 
is  greatest  when  the  connecting  rod  is  at  its  maximum  angle 
with  the  line  of  the  piston  rod.  Multiplying  the  area  of  the 
piston  by  the  average  j)ressure  gives  the  horizontal  push  on 
the  crosshead.  To  tind  the  approximate  maximum  thrust  on 
the  guides,  multiply  this  horizontal  push  by  the  quotient 
obtained  by  dividing  the  length  of  the  crank,  in  inches,  by 
the  length  of  the  connecting  rod,  in  inches. 

Example.  —  Find  the  maximum  thrust  at  right  angles  of  an 
engine  12"  x  24"  with  a  40  lb.  average  pressure  on  the  piston, 
the  length  of  the  rod  being  60"  and  of  the  crank  12". 

Area  of  piston  =  12"  x  12"  x  .7854  =  113.097(5  sq.  in.  or  113  approx. 
113.0976  sq.  in.  x  40  =  4523.9  lb.  horizontal  push  on  crosshead 


4523.9  X  n 


=  905  lb.     Ans. 


Horizontal  Rnoinb 


Weight  of  Flywheel.  —  The  weight  of  the  flywheel  of  an 
engine  may  be  calculated  by  multiplying  the  area  of  the  piston 
by  the  length  of  one  stroke  in  feet,  and  then  multiplying  the 
product  by  the  constant  12,000,000.     Call  thia  product  No.  1. 


224 


VOCATIONAL  MATHEMATICS 


Then  multiply  the  square  of  the  number  of  revolutions  by  the 
square  of  the  diameter  of  the  wheel  in  feet.  Call  this  product 
No.  2.  Divide  product  No.  1  by  product  No.  2,  and  the  quo- 
tient will  give  the  proper  weight  of  the  flywheel  in  pounds. 

Example.  —  Find  the  weight  of  a  flywheel  for  an  engine 
12''  X  24",  if  the  diameter  of  the  wheel  is  6',  and  makes  140 
R.  P.  M. 

12  X  12  X  .7854  =  113.0976  sq.  in.,  ayea  of  piston 
113.0976  X  2  =  226.1952  sq.  in. 
226.1952  X  12,000,000  =  2,714,342,400  (1) 

140  X  140  =  19,600 
19,600  x6x6  =  705,600  (2) 

,  2,714,342,400  --  705,600  =  3847  lb.     Ans. 

Steam  Lap.  —  The  amount  of  steam  lap  that  should  be  added 
to  a  common  slide  valve  is 
obtained  as  follows :  Divide 
the  cut-off  distance  by  the 
distance  of  the  pull  stroke 
and  extract  the  square  root 
of  the  quotient.  Multiply 
this  square  root  by  one  half 
of  the  valve  travel  and  sub- 
tract one  half  of  the  lead 
from  this  product. 

Example.  —  An  engine  has  a  48"  stroke,  with  a  valve  travel 
of  6"  and  a  cut-off  at  half  stroke  and  a  valve  of  I"  lead.  What 
is  the  necessary  steam  lap  ? 


Steam  Chest 
V,  slide  valve        p,  piston 
?•',  valve  rod          r,  piston  rod 


If  =  .5  V.5 

.707x3=2.121    2.121-^ 


.707 

2.121  -  .1250 


1.996".    Ans. 


Horse  Power 

The  power  of  a  steam  engine  is  commonly  designated  as 
horse  power.  By  one  horse  power  is  meant  a  force  great 
enough  to  raise  33,000  lb.  one  foot  high  in  one  minute.     There 


ENGINES  225 

are  two  measures  of  horse  power  in  engines:  indicated,  and 
actual  or  net.  Indicated  horse  power  is  obtained  by  multiply- 
ing the  area  of  the  piston  in  square  inches,  and  the  mean  eifec- 
tive  pressure  in  the  cylinder  in  pounds  per  square  inch,  and  the 
speed  in  feet  per  minute,  and  dividing  the  product  by  33,000. 
The  actual  or  net  horse  power  is  the  difference  between  the 
indicated  horse  power  and  the  amount  expended  in  overcom- 
ing friction. 

Example.  —  What  is  the  horse  power  of  an  engine  which 
can  pump  in  one  minute  68  cu.  ft.  of  water  from  a  depth  of 
108  ft.  ? 

68  X  62^  =  4250  lb.     4250  x  108  =  459,000  ft.  lb. 

MM!^  =  I^  =  13t?  H.  P.    Approx.  14  H.  P.    Ans. 
33,000        11  ^' 

The  horse  power  of  an  engine  is  expressed  by  the  following 
formula : 

A  =  area  of  piston  in  square  inches 

P  =  mean  effective  pressure  of  steam  on  piston  per  sq.  in. 
V  =  velocity  of  piston  per  minute  in  feet 
Ax  PxV 


H.  P.  = 


33,000 


Note.  — A  quick  method  to  find  the  H.  P.  of  an  engine  is  to  square 
the  diameter  of  the  cylinder  in  inches  and  divide  the  product  by  2.5. 
The  quotient  is  approximately  the  H.  P. 

H.P.=^ 
2.5 


D  =  VO  X  H.  f.  -  1.58  VH.  p.     approx. 

Diameter  of  Cylinder.  —  To  find  the  diameter  of  a  cylinder  of 
an  engine  of  a  required  nominal  horse  power: 

PV  .7854 


•>.4- 


7854 


226  VOCATIONAL  MATHEMATICS 

Diameter  of  Supply  Pipe.  —  The  diameter  of  the  steam  supply- 
pipe  for  a  given  engine  may  be  calculated  from  the  H.  P.  of 
the  engine  by  dividing  the  H.  P.  by  6,  and  extracting  the 
square  root  of  the  quotient. 

hTp. 


0=4 


6 

Example.  —  What  is  the  diameter  of  a  steam  supply  pipe 
of  a  216  H.  P.  engine.? 

216  -  6  =  36 

\/36  =  6",  diameter  of  supply  pipe.    Ans. 

EXAMPLES 

1.  Find  the  diameter  of  the  steam  supply  pipe  of  a  180 
H.  P.  engine. 

2.  What  is  the  H.  P.  of  an  engine  whose  area  of  piston  is 
114  sq.  in.,  mean  effective  pressure  80,  and  velocity  of  piston 
112  ft.  per  minute  ? 

3.  What  is  the  approximate  diameter  of  a  cylinder  of  an 
engine  of  50  H.  P.  ? 

4.  What  is  the  approximate  H.  P.  of  an  engine  the  cylinder 
diameter  of  which  is  28"  ? 

5.  What  is  the  effective  area,  for  power  calculation,  of  the 
piston  of  a  steam  engine,  the  bore  of  the  cylinder  being  28", 
and  the  diameter  of  the  piston  rod  which  passes  through  both 
ends  of  the  cylinder  2"  ? 

6.  What  is  the  H.  P.  of  an  engine  that  can  raise  3  tons  of 
coal  (1  ton  =  2240  lb.)  from  a  mine  289  ft.  deep  in  9  minutes? 

7.  What  is  the  approximate  H.  P.  of  an  engine  the  cylinder 
diameter  of  which  is  16"  ? 

8.  Find  the  diameter  of  the  steam  supply  pipe  of  a  24  H.  P. 


ENGINES  227 

9.    What  is  the  approximate  diameter  of  a  cylinder  of  an 
engine  of  80  H.  P.  ? 

10.  What  is  the  approximate  H.  P.  of  an  engine  the  cylinder 
diameter  of  which  is  20"  ? 

11.  What  is  the  H.  P.  of  an  engine  whose  diameter  of  piston 
is  15",  mean  effective  pressure  110,  velocity  of  piston  per 
minute  189'  ? 

12.  How  many  pounds  of  water  per  half  minute  can  an  8  H.  P. 
fire  pump  raise  to  a  height  of  86  ft.  ? 

13.  What  is  the  effective  area  in  square  inches  of  the  piston 
of  a  steam  engine  if  the  diameter  of  the  cylinder  is  20"  and 
the  diameter  of  the  piston  rod  is  3"  ? 

14.  What  is  the  horse  power  of  an  engine  that  is  required 
to  pump  out  a  basement  51'  x  22'  x  10'  deep,  full  of  water,  in 
20  minutes  ? 

15.  (a)  Find  the  diameter  of  the  steam  supply  pipe  of  a 
98  H.  P.  engine. 

(6)  Find  .the  approximate  diameter  of  a  cylinder  of  an  eugine 
of  48  H.  P. 

(c)  What  is  the  approximate  H.P.  of  an  engine  the  cylinder 
diameter  of  which  is  28"  ? 

Steam  Indicator 

In  order  to  know  the  condition  of  the  steam  within  the 
cylinder  of  an  engine  an  indicator  is  used.  This  consists  of 
a  small  cylinder  containing  a  piston,  the  rod  of  which  is  en- 
closed in  a  spiral  spring  which  opposes  the  motion  of  the  piston. 
The  piston  rod,  after  i)assing  through  the  top  of  the  cylinder 
cover,  is  connected  with  a  long  light  lever,  on  the  end  of  which 
is  a  pencil.  This  pencil  moves  in  a  vertical  straight  line  when- 
ever the  piston  moves. 

Another  cylinder  with  an  axis  parallel  to  the  first  carries  a 
paper  drum,  and  this  drum  is  connected  to  the  crosshead  of 
the  engine  by  means  of  a  cord  and  a  reducing  motion,  so  that 


228 


VOCATIONAL  MATHEMATICS 


the  movement  of  the  drum  is  proportional  to  that  of  the  cross- 
head.  When  the  pipe  between  the  indicator  and  the  engine 
is  closed  by  means  of  a  cock,  the  pencil,  when  held  against  the 
drum,  makes  a  horizontal  line  called  an  atmospheric  line.     If 


Cross  Section  of 
Steam  Indicator 


the  cock  is  opened,  admitting  to  the  small  cylinder  of  the  in- 
dicator the  pressure  that  exists  in  the  engine  cylinder,  the  pen- 
cil will  trace  a  figure,  every  point  of  which  is  at  a  height  from 
the  atmospheric  line  proportional  to  the  number  of  pounds' 
pressure  in  the  engine  cylinder  at  every  point  in  the  stroke. 


ENGINES 


229 


Operating  Power. — To  find  the  power  required  to  drive  a 
certain  machine  when  driven  direct  from  the  shaft  or  engine : 
Indicate  the  engine  with  the  machinery  running  and  calcu- 
late from  the  card.  Then  indicate  the  engine  (from  formula 
on  p.  225)  without  the  machinery  running  and  from  this  obtain 
the  H.  P.     The  difference  will  give  the  power  for  operating. 

To  indicate  an  engine  means  to  place  a  steam  engine  indicator  on  the 
steam  engine  and  record  on  the  diagrams  the  pressure  in  the  steam  cyl- 
inder at  every  part  of  the  stroke.  The  diagrams  are  measured  by  means 
of  a  planimeter  or  by  means  of  ordinutes.  The  latter  way  may  be  done 
by  dividing  the  diagram  into  10  or  20  equal  spaces  by  vertical  lines. 
On  these  verticals  measure  the  length  between  the  back  pressure  line  and 
mark  each  length  on  a  long  strip  of  paper.  Dividing  the  sum  of  all  these 
lengths  by  10  or  20  will  give  the  average  length  of  ordinate  which  multiplied 
by  the  scale  of  the  spring  will  give  the  mean  effective  pressure  (M.E.P.). 


Diagram  from  Hartford  Engine 
Cylinder,  IH  X  24  inches.     B<»iler  pressure,  87  pounds.    Vacuum  by  gauge,  23i 
inches.    130  revolutions  per  minute.    Scale,  50.     The  vertii-al  lines  from  A 
are  called  ordinates. 

Example.  —  If  the  total  length  is  9"  and  the  spring  used 
is  40  lb.,  how  is  the  mean  effective  pressure  (M.  E.  P.)  found  ? 

40  X  .9  =  36  lb.  M.  E.  P.    Ans. 


230  VOCATIONAL  MATHEMATICS 

EXAMPLES  1 

1.  If  the  total  length  of  the  ordinates  is  11"  and  the  spring 
used  is  56  lb.,  what  is  the  M.  E.  P.  ? 

2.  A  9  X  10  engine  has  a  48  lb.  average  pressure  on  the 
piston,  the  length  of  rod  is  35",  and  the  crank  is  5".  Find  the 
maximum  thrust  at  right  angles. 

3.  Find  the  weight  of  a  flywheel  of  an  engine  12''  x  24", 
diameter  of  wheel  7',  with  168  K.  P.  M. 

4.  What  is  the  necessary  steam  lap  of  an  engine  with  a 
45"  stroke,  valve  travel  5",  and  cut-off  at  half  stroke,  and  valve 
i"  lead  ? 

5.  If  the  total  length  of  ordinates  is  10"  and  the  spring 
used  is  49  lb.,  what  is  the  M.  E.  P.  ? 

6.  A  12''  X  24"  engine  has  a  50  lb.  average  pressure  on  the 
piston,  the  length  of  the  rod  is  59",  and  the  crank  is  12" ;  what 
is  the  maximum  thrust  at  right  angles  ? 

7.  What  is  the  necessary  steam  lap  of  an  engine  with  a 
64"  stroke,  valve  travel  7,  cut-off  at  half  stroke,  and  valve  ^" 
lead  ? 

8.  If  the  total  length  of  ordinates  is  8"  and  the  spring  used 
is  38  lb.,  what  is  the  M.  E.  P.  ? 

9.  Find  the  weight  of  a  flywheel  of  an  engine  24"  x  60", 
the  diameter  of  wheel  being  24'  and  having  75  R.  P.  M. 

10.  If  the  total  length  of  ordinates  is  12"  and  the  spring 
used  is  61  lb.,  what  is  the  M.  E.  P.  ? 

11.  Find  the  weight  of  a  flywheel  of  an  engine  30"  x  60", 
the  diameter  of  the  wheel  30',  with  63  K  P.  M. 

12.  An  engine  10"  x  12"  has  a  35  lb.  average  pressure  on  the 
piston,  the  length  of  the  rod  is  42",  and  the  crank  is  6".  Find 
the  maximum  thrust  at  right  angles. 

1  Use  10  ordinates  in  solving  problems. 


PART  VIII  —  MATHEMATICS    FOR   ELECTRICAL 
WORK 


CHAPTER   XV 
COMMERCIAL  ELECTRICITY 

Amperes. — What  electricity  is  no  one  knows.  Its  action, 
however,  is  so  like  that  of  flowing  water  that  the  comparison 
is  helpful.  A  current  of  water  in  a  pipe  is  measured  by  the 
amount  which  flows  through  the  pipe  in  a  second  of  time,  as 
one  gallon  per  second.     So  a  current  of  electricity  is  measured 


Water  Analogy  of  Fall  of  Potential 

by  the  amount  which  flows  along  a  wire  in  a  second,  as  one 
coulomb  per  second,  —  a  coulomb  being  a  unit  of  measurement 
of  electricity,  just  as  a  gallon  is  a  unit  of  measurement  of 
water.  The  rate  of  flow  of  one  coulomb  per  second  is  called  one 
ampere.  The  rate  of  flow  of  five  coulombs  per  second  is  five 
amperes. 

Volts.  —  The  quantity  of  water  which  flows  through  a  pipe 
depends  to  a  large  extent  upon  the  pressure  under  which  it 
flows.     The  number  of  amperes  of  electricity  which  flow  along 

231 


232  VOCATIONAL  MATHEMATICS 

a  wire  depends  in  the  same  way  upon  the  pressure  behind  it. 
The  electrical  unit  of  pressure  is  the  volt.  In  a  stream  of 
water  there  is  a  difference  in  pressure  between  a  point  on  the 
surface  of  the  stream  and  a  point  near  the  bottom.  This  is 
called  the  difference  or  drop  in  level  between  the  two  points. 
It  is  also  spoken  of  as  the  pressure  head,  "  head  "  meaning  the 
difference  in  intensity  of  pressure  between  two  points  in  a  body 
of  water,  as  well  as  the  intensity  of  pressure  at  any  point. 
Similarly  the  pressure  (or  voltage)  between  two  points  in  an 
electric  circuit  is  called  the  difference  or  drop  in  pressure  or 
the  poteyitial.  The  amperes  represent  the  amount  of  electricity 
flowing  through  a  circuit,  and  the  volts  the  pressure  causing 
the  flow. 

Ohms.  —  Besides  the  pressure  the  resistance  of  the  wire 
helps  to  determine  the  amount  of  the  current: — the  greater 
the  resistance,  the  less  the  current  flowing  under  the  same 
pressure.  To  the  electrical  unit  of  resistance  the  name  ohm 
is  given.  A  wire  has  a  resistance  of  one  ohm  when  a  pressure 
of  one  volt  can  force  no  more  than  a  current  of  one  ampere 
through  it. 

Ohm's  Law.  —  The  relation  between  current  (amperes), 
pressure  (volts),  and  resistance  (ohms)  is  expressed  by  a  law 
known  as  Ohm^s  Law.  This  is  the  fundamental  law  of  the 
study  of  electricity  and  may  be  stated  as  follows : 

An  electric  current  flowing  along  a  conductor  is  equal  to 
the  pressure  divided  by  the  resistance. 

Current  (amperes)  =  - — — ^— — ^ 

Resistance  (ohms) 

Letting  /=  amperes, -EJ  =  volts,  i?  =  ohms, 

I=E-r-  Rov  I  =  ~ 
R 

E=IR 

R  =  ^ 
I 


ELECTRICAL  WORK  233 

Example.  —  If  a  pressure  of  110  volts  is  applied  to  a  re- 
sistance of  220  ohms,  what  current  will  flow  ? 

/  =  ^  =  li2  =  1  =  .6  ami^re.     Ans. 
R     220     2 

Example.  —  A  current  of  2  amperes  flows  in  a  circuit  the 
resistance  of  which  is  300  ohms.     ^Vhat  is  the  voltage  of  the 

circuit  ? 

IIi  =  E 
2  X  300  =  600  volte.     Ans. 

Example.  —  If  a  current  of  12  amperes  flows  in  a  circuit 
and  the  voltage  applied  to  the  circuit  is  240  volts,  find  the 
ijesistance  of  the  circuit. 

^=Ji        ?i2  =  20  ohms.     Ans. 

Ammeter  and  Voltmeter.  —  Ohm's  Law  may  be  applied  to  a 
circuit  as  a  whole  or  to  any  part  of  it.     It  is  often  desirable  to 


/ 


Ammbtkb  Voltmetke 

know  how  much  current  is  flowing  in  a  circuit  without  calcu- 
lating it  by  Ohm's  Law.  An  instrument  called  an  ammeter  is 
used  to  measure  the  current.  This  instrument  has  a  low 
resistance  so  that  it  will  not  cause  a  drop  in  pressure.  A 
voltmeter  is  used  to  measure  the  voltage.  This  instrument  has 
high  resistance  so  that  a  very  small  current  will  flow  through 


234  VOCATIONAL  MATHEMATICS 

it,  and  is  always  placed  in  shunt,  or  parallel  (see  p.  235)  with 
that  part  of  the  circuit  the  voltage  of  which  is  to  be  found. 

Example.  —  What  is  the  resistance  of  wires  that  are  carry- 
ing 100  amperes  from  a  generator  to  a  motor,  if  the  drop  or 
loss  of  potential  equals  12  volts  ? 


Drop  in  voltage  =  IB 

I  =  100  amperes 

Drop  in  volts      =  12 

B  =      ?  ohms 

-f 

B- 

=  —  =  0.12  ohm.     J 
100 

Ans. 

Example.  —  A  circuit  made  up  of  incandescent  lamps  and 
conducting  wires  is  supplied  under  a  pressure  of  115  volts. 
The  lamps  require  a  pressure  of  110  volts  at  their  termiiials 

+  lead 


Wiring  of  Incandescent  Lamp  Circuit 

and  take  a  current  of  10  amperes.  What  should  be  the  resist- 
ance of  the  conducting  wires  in  order  that  the  necessary  cur- 
rent may  flow  ? 

Drop  in  conducting  wireS  =  115  —  110  =  5  volts 
Current  through  wires       =    10  amperes 

E      5 
B  =  —  zz  —  =  0.5  ohm  resistance.     Ans. 
I      10 

EXAMPLES 

1.  How  much  current  will  flow  through  an  electromagnet 
of  140  ohms'  resistance  when  placed  across  a  100-volt  circuit  ? 

2.  How  many  amperes  will  flow  through  a  110-volt  lamp 
which  has  a  resistance  of  120  ohms  ? 

3.  What  will   be  the   resistance   of   an   arc   lamp  burning 
upon  a  110-volt  circuit,  if  the  current  is  5  amperes  ? 


ELECTRICAL  WORK 


235 


4.  If  the  lamp  in  Example  3  were  to  be  put  upon  a  150- 
volt  circuit,  how  much  additional  resistance  would  have  to  be 
put  into  it  in  order  that  it  might  not  take  more  than  5 
amperes  ? 


Motor 

Electric  Road  System 

5.  In  a  series  motor  used  to  drive  a  street  car  the  resistance 
of  the  field  equals  1.06  ohms;  the  current  going  througli  equals 
30  amperes.  What  would  a  voltmeter  indicate  if  placed 
across  the  field  terminals  ? 

6.  If  the  load  upon  the  motor  in  Example  5  were  increased 
so  that  45  amperes  were  flowing  through  the  field  coils,  what 
would  the  voltmeter  then  indicate  ? 


Series  and  Parallel  Circuits 

Pieces  of  electrical  apparatus  may  be  connected  in  two  ways. 
When  the  pieces  are  connected  so  that  the  current  passes 
through  them  in  a  single  path,  they  are  said  to  be  in  series. 


Wmf\  - 

B 

Cells  Connected  in  Parallel 


^}\m\ 


Cells  Connected  in  Series 


WTien  the  pieces  are  connected  so  that  the  current  is  divided 

between  them,  they  are  said  to  be  in  jxtrallel  with  one  another.. 

The  total  resistance  of  a  series  circuit  is  equal  to  the  sum  of 

the  resistances  of  the  separate  parts  of  the  circuit.    The  total 


236  VOCATIONAL  MATHEMATICS 

voltage  of  a  series  circuit  is  equal  to  the  sum  of  the  voltages 

across  the  separate  resistances. 

A /?i /?2 B 

Total  resistance  A  to  B  =  Ei  -\-  R.^- 

Example.  —  If  there  is  a  circuitof  240  volts  and  the  lamps 
are  of  240  ohms'  resistance  but  made  to  carry  only  ^  an  ampere, 
two  lamps  would  have  to  be  put  in  series  in  order  to  use  them 
on  the  240-volt  circuit. 

The  resistance  of  the  two  lamps  in  series  would  then  be  480  ohms, 
the  voltage   of  the   circuit  240  volts,  and  the   current  by   Ohm's  Law 

I=  —  =  —  =  -  ampere. 
H      480     2 

A i?i R-2 . B 

In  any  closed  circuit,  the  algebraic  sum  of  the  products 
found  by  multiplying  the  resistance  of  each  part  by  the  current 
passing  through  it,  is  equal  to  the  voltage  of  the  circuit. 

Example.  —  If  three  lamps  of  110  ohms'  resistance  are  con- 
nected in  series  and  take  i  ampere,  the  voltage  of  the  circuit 

IiBi  4-  IoIi2  +  Is^i  = 

(^x  110)  +  (i  X  110)  +  (^x  110)  = 
55  +  55  -f  55  =:  165  volts.     Ans. 

Example.  —  A  current   of   50   amperes   flowed   through   a 

circuit  when  the  voltage  was  550.     What  resistance  should  be 

added  in  series  with  the  circuit  to  reduce  the  current  to  11 

amperes  ?       _    . 

Resistance  =  ^■^^-  =  11  ohms 

Resistance  =  ^f^-  =  50  ohms 

Additional  resistance  =  50  —  11  =  39  ohms.     Ans. 

Example.  —  The  voltage  required  by  15'  arc  lamps  connected 
in  series  is  900  and  the  current  is  6  amperes.  If  the  resistance 
of  the  connecting  wires  is  5  ohms,  how  much  additional  volt- 
age will  be  necessary  so  that  the  lamp  voltage  may  not  drop 
below  900  ? 

Drop  in  voltage  in  connecting  wires  =  E  =  IE 

6  X  5  =  30  volts  =  additional  voltage  necessary.     Ans. 


ELECTRICAL  WORK  237 

Example. — The  field  coil  of  a  motor  having  4  poles  is 
measured  for  voltage  across  the  terminals  and  the  following 
readings  are  taken  : 

Voltage  across  line  =  220 

Voltage  across  coil  No.  1  =  78.33 
Voltage  across  coU  No.  2  =  00.00 
Voltage  across  coil  No.  3  =  73.33 
Voltage  across  coil  No.  4  =  73.33 
Current  flowing  =1.5  amperes. 

What  is  the  total  resistance  and  what  is  the  trouble  at  coil 
No.  2  ? 

Total  resistance  =  7?  =  ^  =  ^=^  =  140.6  ohms.     Am. 
I      1.5 

As  thei-e  is  no  drop  across  the  terminal  of  coil  No.  2,  there  is  prac- 
tically no  resistance  and  the  current  is  not  going  around  the  coil  but 
throusjli  a  path  of  extremely  low  resistance. 

EXAMPLES 

1.  If  three  electromagnets  are  connected  in  series  and  the 
resistances  are  3,  5,  and  17  ohms,  respectively,  what  is  the 
total  resistance  of  this  set  ? 

2.  The  field  coils  of  a  series  motor  have  a  resistance  of  10 
ohms  and  the  armature  has  a  resistance  of  7  ohms ;  what  is  the 
total  resistance  of  the  motor  ? 

3.  («)  What  would  be  the  total  resistance  of  two  110-volt 
incandescent  lamps  placed  in  series  across  a  110-volt  line  if 
each  lamp  has  a  resistance  of  220  ohms?  {h)  Would  these 
lamps  light  on  this  voltage  in  this  position  ?     (c)  Why  ? 

4.  Three  coils  are  connected  in  series  and  have  a  resistance 
of  3,  5,  and  8  ohms,  respectively.  What  current  will  flow  if 
the  voltage  of  the  circuit  is  64  ? 

5.  Five  arc  lamps,  each  having  a  resistance  of  4  ohms,  are 
connected  in  series.  The  resistance  of  the  connecting  wires 
and  the  other  apparatus  is  o  ohms.  What  must  be  the  voltage 
of  the  circuit  so  that  a  current  of  10  amperes  may  flow? 


238  VOCATIONAL  MATHEMATICS 

6.  A  current  of  10  amperes  was  passing  througli  a  circuit 
under  a  pressure  of  550  volts.  The  circuit  was  made  up  of 
three  sections  connected  in  series,  and  the  resistance  of  two 
sections  was  8  and  12  ohms,  respectively.  What  was  the 
resistance  of  the  third  section  ? 

Example.  —  If  5  electromagnets  are  arranged  in  series  and 
marked  A,  B,  C,  D,  and  E,  the  resistance  of  the  circuit  is  45 
ohms  and  the  resistance  of  each  coil  is  :  ^,  5  ohms ;  B,  10  ohms  ; 
C,  7  ohms;  D,  8  ohms;  E,  15  ohms.  How  much  E.  M.  F. 
would  be  required  to  cause  10  amperes  to  flow  through  the 
coils,  and  what  would  be  the  E.  M.  F.  across  the  terminals  of 

each  coil  ? 

Voltage  across  ^  =  50 
Voltage  across  jB  =  100 
Voltage  across  C  =  70 
Voltage  across  i>  =  80 
Voltage  across  E  =  150 
450 
10  X  45  =  450,  total  voltage.    Ans. 

Example.  —  What  E.  M.  F.  is  necessary  to  send  a  current 
through  10  field  coils  connected  in  series,  if  each  has  a  resist- 
ance of  10  ohms  and  3  amperes  are  required  to  produce  the 
necessary  magnetization  ? 

10  ohms  =  resistance  of  each  coil 

10  coils  are  in  series 

10  X  10  =  100  ohms,  total  resistance 

B  =  100  ohms 

7=3  amperes 

E  =  IB  =3x100=  300  volts.     Ans. 

In  any  closed  circuit  the  algebraic  sum  of  the  products  found 
by  multiplying  the  resistance  of  each  part  by  the  current 
passing  through  it,  is  equal  to  the  voltage  of  the  circuit.  This 
is  practically  an  inverse  statement  of  the  law  of  series  circuits. 

Example.  —  If  we  have  five  lamps  of  110  ohms'  resistance, 
connected  in  series  and  taking  ^  ampere,  the  voltage  of  the 
circuit  is : 


ELECTRICAL  WORK  239 

E  =  IxRi  +  hRi  +  7,^3  +  URi  +  /»/?»  =  i  X  110  +  i  X  110  -f  i  X  110 
+  \  X  110  +  i  X  110  =  66  +  66  +  66  -h  65  -f-  66  =  276  volts.     Ans. 

In  a  parallel  circuit  the  voltage  across  each  branch  is  the 
same  as  the  voltage  across  the  combination.  The  current  is 
equal  to  the  sum  of  the  currents  in  the  separate  parts. 

The  resistance  is  equal  to  the  reciprocal  of  the  sum  of  the 
conductances  of  the  separate  parts. 

Conductance  is  the  reciprocal  of  resistance  and  is  equal  to 
— .     The  unit  of  conductance  is  mho  (ohm  spelled  backwards). 

mho=  — —  or  ohm  = 

ohm  mho 

Series  and  parallel  circuits  may  be  combined  and  may  exist 
in  the  same  circuit.  In  parallel  circuits  the  reciprocal  of  the 
total  resistance  is  equal  to  the  sum  of  the  reciprocals  of  the 
paralleled  resistance. 

If  Rq  =  resistance  of  circuit  and 
rj,  r2,  and  r^  =  parallel  resistances, 

ri _ 

A r^ B 

rg 

resistance  between  A  and  B  =  Rq. 

R     n      r^      'i\ 
Example.  —  Suppose   that  ?•,,  Vz,    and  r^  are  lamps  of  300 
ohms'  resistance  each,  then 

Bo     300     300     300     iWO      100 
Bo  =  100  ohms.     Ans. 
Or,  ri  =  100  ohms  J__J_   ,_L,_L_  11L__L 

rj=    60  ohms  Bo      100      60     300~300~30 

rs  =  300  ohms  Bo  =  30  ohms.     Ans. 

When  it  is  necessary  to  get  tlie  resistance  of  parallel 
circuits,  it  is  often  more  convenient  to  use  the  sum  of  the 
conductances  as  the  total  conductance  of  the  circuit. 


240  VOCATIONAL  MATHEMATICS 

Example.  —  Three  resistances  in  parallel : 
7*1  =  2  ohms 


^2  =  5  ohms 

rg  =  1  olim 

A 

I'n 

Conductance 

»  H- 

.5  mho 

<^>H' 

:    .2  mho 

<"  rr 

:  1     mho 
1.7  mho'conductanc 

Resistance  of  circuit  AB  =  — -  =  .58^ 

Ans. 


EXAMPLES 

1.  Ten  arc  lamps  of  200  ohms'  resistance  are  connected  in 
series  and  the  voltage  of  the  circuit  is  300.  How  much  current 
does  each  lamp  take  ? 

2.  What  will  be  the  E.  M.  F.  necessary  to  supply  60  Thomp- 
son-Houston arc  lamps  arranged  in  series,  the  resistance  of  each 
lamp  being  5  ohms  when  burning,  making  a  total  resistance  of 
300  ohms  in  the  circuit,  if  the  current  required  is  10  amperes  ? 

3.  Four  parallel  circuits  of  2,  4,  5,  and  10  ohms'  resistance, 
respectively,  have  40  volts  impressed  upon  their  terminals, 
(a)  What  is  the  total  current  flowing  ?  (b)  How  much  current 
flows  in  each  branch  ? 

4.  Three  incandescent  lamps  of  different  sizes  are  placed  in 
parallel  on  a  circuit.  These  have  respective  resistances  of  100, 
150,  and  300  ohms.  What  is  the  total  current  flowing  through 
these  lamps  when  the  pressure  applied  is  100  volts? 


ELECTRICAL  WORK  241 

5.  What  is  the  total  resistance  of  four  incandescent  lamps 
placed  in  parallel,  if  each  lamp  has  a  resistance  of  220  ohms  ? 

6.  What  is  the  resistance  of  a  shunt-wound  generator,  if  the 
field  and  the  armature  are  respectively  25  and  10  ohms  ? 

7.  Two  coils  on  a  large  electromagnet  for  lifting  iron  ore  are 
connected  in  parallel  and  the  resistance  of  each  coil  is  40  ohms. 
What  is  the  whole  resistance  ? 

8.  The  resistance  of  a  line  frou)  a  power  house  to  a  mill  is 
6  ohms;  there  are  50  lamps  in  the  mill,  each  lamp  having  a  re- 
sistance of  220  ohms.  What  is  the  total  resistance  from  the 
power  house  ? 

9.  In  an  electric  street  car  4  heaters  are  all  connected  in 
series  and  each  has  a  resistance  of  20  ohms  with  a  voltage  of 
600  across  the  circuit,  (a)  What  is  the  total  resistance  of 
these  ?     (b)  How  many  amperes  will  go  through  them  ? 

Power  Measurement 
The  flow  of  an  electric  current  has  been  compared  to  the  flow 
of  water  through  a  pipe.  The  water  current  is  measured  by 
the  number  of  gallons  or  pounds  flowing  per  minute.  A  cur- 
rent of  electricity  is  measured  by  the  number  of  amperes  or 
coulombs  per  second.  When  a  gallon  of  water  is  raised  a  foot 
by  means  of  a  pump  a  certain  amount  of  work  is  done.  So 
when  a  coulomb  of  electricity  is  passed  through  a  wire  under 
the  pressure  of  one  volt,  a  certain  amount  of  work  is  done.  In 
the  case  of  water  the  work  done  is  measured  in  foot  pounds. 
A  foot  pound  is  the  work  done  in  raising  a  weight  of  one  pound 
through  a  distance  of  one  foot. 

Work  =  Force  x  Distance 

When  one  coulomb  of  electricity  is  passed  through  a  wire 
under  a  pressure  of  one  volt,  the  amount  of  work  done  is  called 
one  joule. 

The  ix)wer  required  to  keep  a  current  of  water  flowing  is  the 
product  of  the  current  in  pounds  per  minute  by  distance  in  feet. 


242 


VOCATIONAL  MATHEMATICS 


This  gives  the  power  in  foot  pounds  per  minute.  Mechanical 
power  is  usually  expressed  in  horse  power  (H.  P.).  The  power 
required  to  keep  a  current  of  electricity  flowing  is  the  product 
of  the  current  in  amperes  by  the  pressure  in  volts  and  is  ex- 
pressed in  watts. 

1  H.  P.  =  746  watts 
1000  watts  =  1  kilowatt 
1  kilowatt  (K.  W.)  =  1.34  or  li-  H.  P. 
Volts  X  Amperes  ^j^.j^^^^^^ 


lOUO 

Let 

P  =  power  in  watts 

/  =  current  in  amperes 

E  =  pressure  in  volts 

then  (1) 

F=  IE  (equation  for  power) 

but 

E=TB  (Ohm's  Law) 

therefore 

P=  I  (IB)  or  PB  (by  substitution) 

(2) 

P=I^B 

but 

7  =  ^  (Ohm's  Law) 
B 

therefore 

P  =  ^  (E)  or  ^  (by  substitution) 
B                B 

(3) 

■   -f 

thus 


P=  IE=  DB  = 


E^ 
B 


To  measure  power  a  wattmeter  is  used,  which  is  a  combina- 
tion of  a  voltmeter  and  an  ammeter. 

In  order  to  find  the  amount  of  work  done  by  a  certain  engine, 
it  is  necessary  to  know  the  time  it  has  been  running  and  the 
power  it  has  been  supplying,  i.e.  its  rate  of  doing  work.  If 
the  power  is  measured  in  horse  power  and  the  time  in  hours, 
the  work  is  done  in  horse  power  hours.  Similarly,  if  the 
power  is  measured  in  kilowatts  and  the  time  in  hours,  the  work 
done  is  measured  in  kilowatt  hours  (K.  W.  H.). 

1  H.  P.  H.  =  0.746  K.W.  H. 
1  K.  W.  H.  =  1.84    H.  P.  H. 


ELECTRICAL  WORK  243 

These  units  are  too  large  to  be  used  conveniently  in  all  problems, 
so  a  smaller  electrical  unit  called  the  watt  secondf  or  joulCf  is 
used. 

EXAMPLES 

1.  If  the  resistance  of  a  circuit  is  1  ohm  and  the  current  30 
amperes,  what  energy  is  expended  in  one  half  hour  ? 

2.  With  a  potential  difference  of  95  volts  and  a  current  of 
15  amperes,  what  energy  is  expended  in  20  minutes? 

3.  If  a  current  of  100  amperes  flows  for  2  minutes  under  a 
pressure  of  500  volts,  what  is  the  work  done  in  joules  ? 

4.  If  12  incandescent  lamps  burn  for  10  hours  under  a 
pressure  of  110  volts,  each  lamp  consuming  J  an  ampere,  how 
many  kilowatt  hours  are  used  ? 

5.  Fifty  horse  power  expended  continuously  for  one  hour 
will  produce  how  many  kilowatt  hours  ? 

6.  If  4000  watts  are  expended  in  a  circuit,  how  much  horse 
power  is  being  developed  ? 

7.  If  20  horse  power  of  mechanical  energy  were  converted 
into  electrical  energy,  how  many  watts  would  be  developed  ? 

8.  If  a  current  of  50  amperes  flows  through  a  circuit  under 
a  pressure  of  220  volts,  what  is  the  power  ? 

9.  If  200  watts  are  expended  in  a  circuit  by  a  current  of  4 
amperes,  what  is  the  voltage  required  to  drive  the  current 
through  the  wire  ? 

10.  If  an  incandescent  lamp  requires  \  an  ampere  of  current 
and  the  resistance  of  its  filament  is  220  ohms,  how  many  watts 
are  required  for  it  ? 

Measurement  of  Resistance 

The  amount  of  current  in  a  circuit  depends  upon  the  voltage 
and  upon  the  resistance.  To  control  the  current  it  is  necessary 
to  change  one  of  these  two  factors.  The  resistance  to  the  flow 
of  water  through  a  pipe  depends  upon  the  shape  of  the  pipe  and 


244  VOCATIONAL  MATHEMATICS 

its  length.  The  electrical  resistance  of  a  conductor  depends 
upon  the  nature  of  the  metal  from  which  the  conductor  is  made, 
its  size,  its  length,  and  the  temperature.  The  greater  the  size 
of  the  conductor,  the  greater  is  its  power  for  conducting  elec- 
tricity, and  therefore  the  less  its  resistance.  The  longer  the 
wire,  the  less  its  conducting  power,  and  therefore  the  greater 
the  resistance.  As  the  resistance  of  a  large  pipe  is  less  than 
the  resistance  of  a  small  one,  so  the  resistance  of  a  large  wire 
is  less  than  the  resistance  of  a  small  one. 

Copper  is  the  material  generally  used  for  wires,  and  its  con- 
ductivity, or  capacity  for  conducting  current,  is  taken  as  the 
standard.  The  conductivity  of  pure  copper  is  expressed  as 
100  % .  Commercial  copper  usually  has  from  98  to  99  %  con- 
ductivity. Other  materials  used  in  electric  wires  are  iron, 
aluminum,  brass,  etc.  Iron  has  a  conductivity  of  about  16  % 
and  brass  of  about  25  %.  It  would  require  an  iron  wire  with 
over  6  times  the  cross  section  of  a  copper  wire  to  give  the  same 
conductivity,  and  brass  wire  would  have  to  be  about  4  times  as 
large  in  its  cross  section  for  the  same  conductivity.  Generally 
it  is  assumed  that  all  electric  wires  are  copper. 

In  measuring  the  length  of  wires  the  unit  used  is  feet,  while 
the  cross  section  area  is  measured  in  circular  mils;  y^Vir  ^^ 
an  inch  is  called  a  m^7,  and  a  round  wire  one  mil  in  diameter 
is  said  to  have  a  cross-section  area  of  one  circular  mil.  A 
wire  1  foot  long,  with  a  cross-section  area  of  1  circular  mil, 
is  called  a  mil-foot  wire.  The  area  of  any  wire  in  circular  mils 
may  be  found  by  squaring  the  number  of  thousandths  of  an  inch 
in  the  diameter. 

If  B  =  resistance  of  wire  in  ohms 

L  =  resistance  of  wire 
D  =  diameter  of  wire  in  mils 
cP  =  area  of  wire  in  circular  mils 
K  =  resistance  of  1  mil  foot  in  ohms  called 

resistivity  or  specific  resistance  of  material 

then  B  =  ^ 


ELECTRICAL  WORK  245 

Note.  —  /T  varies  with  the  niAterial  and  the  temperature.  The  resist- 
ance of  1  mil  foot  of  soft  copper  wire  at  60°  F.  is  10.4  ohms. 

Example.  —  What  is  the  area  of  a  wire  0.1  inch  in  diameter  ? 

0.1  inch  =  100  thouBandths  of  an  inch 
100  X  100  =  10,000,  number  of  circular  ihils.     Ans. 

Example.  — What  is  the  resistance  at  ordinary  temi)erature 
of  a  copper  wire  2500  ft.  long  with  a  cross-section  area  of 
10,000  circular  mils  ?     (K=  10.) 

^^^^10^^^500^2.6  ohms.    Aus. 
d^  10000 


EXAMPLES 

1.  What  must  be  the  diameter  of  a  wire  in  mils  in  order 
that  it  may  have  a  cross-section  area  of  200  circular  mils  ? 

2.  How  many  circular  mils  are  there  in  a  wire  50  mils  in 
diameter  ? 

3.  How  many  circular  mils  are  there  in  a  wire  150  mils  in 
diameter  ? 

4.  What  is  the  diameter  in  inches  of  a  copper  wire  which 
has  a  cross-section  area  of  20,000  circular  mils  ? 

5.  If  it  is  desired  to  have  a  copper  wire  of  J  ohm  resistance 
and  2000  ft.  long,  what  must  its  cross  section  be  ? 

6.  What  pressure  is  required  to  force  a  current  of  50  am- 
peres over  a  copper  wire  IGOO  ft.  long  which  has  a  cross- 
section  area  of  20,000  circular  mils  ? 

7.  A  current  is  forced  through  a  copper  wire  2000  ft.  long 
under  a  pressure  of  50  volts.  If  the  wire  has  an  area  of  5000 
circular  mils,  what  is  the  value  of  the  current  flowing? 

8.  If  a  wire  5000  ft.  long  carries  a  current  of  5  amperes 
under  a  pressure  of  100  volts,  what  is  the  cross-section  area  of 
the  wire  ? 


246  VOCATIONAL  MATHEMATICS 

9.  A  wire  100  ft.  long  is  in  series  with  another  wire  500  ft, 
long  and  the  first  wire  is  ^"  in  diameter.  If  the  second  wire 
has  a  cross-section  area  of  20,000  circular  mils,  what  is  the 
resistance  of  the  circuit  ? 

10.  How  many  amperes  will  a  wire  ^"  in  diameter  carry  if 
a  wire  1000  mils  in  diameter  will  carry  650  amperes  ? 

Size  of  Wire 

If  an  electrician  wishes  to  know  the  size  of  a  wire  to  carry 
a  certain  current  a  certain  distance  with  a  certain  drop  of 
voltage,  he  may  ascertain  it  by  substituting  values  in  the  foL 
lowing  formula,  which  is  called  a  two-wire  formula, 

e 
CM  =  size  of  wire  in  circular  mils 
D  =  distance  from  distribution 
7=  amperage 
e  =  drop  or  volts  lost 

Example.  —  What  size  of  wire  will  be  required  for  a  motor 
situated  85  ft.  from  the  center  of  distribution,  if  the  motor  is 
5  H.  P.,  operating  at  difference  of  potential  of  110  volts,  allow- 
ing 3  %  drop  ? 

85  ft. 


Ans. 


P=ET 
1  H.  P.  =  746  watts 

1=^     j^3730      g:=  no  X. 03=3.30     D= 

P  =  746  X  6  =  3730 

E=no 

21.6x85x^^3^ 

110  _  21.6  X  85  X  3730 

3.30                     3.30  X  110 

=  18865.7  wire. 

REVIEW  EXAMPLES 


1.  What  size  of  wire  will  be  required  for  a  15  H.  P.  motor 
operating  at  550  volts  and  situated  35  ft.  from  the  center  of 
distribution,  allowing  a  2  %  drop  ? 


ELECTRICAL  WORK  247 

2.  J  low  many  coulombs  are  delivered  in  a  minute  when 
the  current  is  17^  amperes  ? 

3.  What  is  the  current  when  480  coulombs  are  delivered 
per  minute  ? 

4.  In  what  time  will  72,000  coulombs  be  delivered  when 
the  current  is  80  amperes? 

5.  What  size  of  wire  will  be  required  for  a  7^  H.  P.  motor 
operating  at  220  volts  and  situated  65  ft.  from  the  center  of 
distribution  and  allowing  a  5  %  drop  ? 

6.  If  a  current  of  20  amperes  flows  through  a  circuit  for 
21.2  hours,  what  quantity  of  electricity  is  delivered? 

7.  How  many  ampere  hours  pass  in  a  circuit  in  2^  hours 
when  the  current  is  1 6  amperes  ? 

8.  What  size  of  wire  will  be  required  for  a  10  H.  P.  motor 
operating  at  110  volts  and  situated  105  ft.  from  the  center  of 
distribution,  allowing  a  drop  of  3  volts? 

9.  If  200  coulombs  of  electricity  are  passed  through  an 
electrolytic  vat  each  second  under  a  pressure  of  10  volts,  how 
many  joules  of  work  are  expended  in  an  hour  ? 

10.  What  quantity  of  electricity  must  flow  under  a  pressure 
of  5  volts  to  do  125  joules  of  work  ? 

11.  If  10  coulombs  do  10  joules  of  work  flowing  through 
a  wire,  what  is  the  pressure  ? 

12.  What  must  be  the  diameter  of  a  wire  in  mils  in  order 
that  it  may  have  a  cross  section  of  200  circular  mils? 

13.  A  wii-e  5000  ft.  long  carries  a  current  of  6  amperes 
under  a  pressure  of  100  volts.  What  is  the  cross-section  area 
of  the  wire  ? 

14.  A  copper  wire  ^  inch  in  diameter  and  100  ft.  long  is  in 
series  with  another  copper  wire  500  ft.  long  with  a  cross 
section  of  20,000  circular  mils.  What  is  the  resistance  of  the 
circuit  ? 


248  VOCATIONAL  MATHEMATICS 

15.  What  is  the  resistance  at  ordinary  temperature  of  a 
copper  wire  2500  ft.  long  having  a  cross-section  area  of  10,000 
circular  mils  ? 

16.  If  it  is  desired  to  have  a  copper  wire  of  i  ohm  resistance 
and  2000  ft.  long,  what  must  be  its  cross-section  area? 

17.  A  voltmeter  which  measures  the  pressure  on  a  circuit 
registers  500  volts  and  the  ammeter  on  the  same  circuit  shows 
25  amperes.     What  is  the  resistance  of  the  circuit  ? 

18.  An  electromagnet  has  a  resistance  of  25  ohms,  and  there 
must  be  41  amperes  passing  through  it  in  order  that  the 
magnetism  may  be  strong  enough.     What  must  be  the  voltage  ? 

19.  What  current  is  needed  to  light  a  16  C.  P.  lamp,  if  the 
hot  resistance  of  the  lamp  is  220  ohms  and  the  voltage  is  110? 

20.  A  storage  battery  gives  2.3  volts  and  it  is  connected 
with  a  coil  having  a  resistance  of  25  ohms.  What  current 
will  flow  through  the  circuit  if  the  internal  resistance  of  the 
cell  is  zero? 

21.  What  is  the  power  necessary  to  drive  a  current  of  500 
amperes  through  a  resistance  of  5  ohms  ? 

22.  How  many  watts  of  power  are  going  to  an  electric  motor 
if  the  voltage  of  the  line  is  500  and  there  are  7  amperes  enter- 
ing the  motor  ? 

23.  How  many  H.  P.  are  needed  to  run  a  dynamo  that  is 
lighting  258  lamps  in  parallel,  if  each  lamp  takes  |  an  ampere 
at  110  volts  ? 

24.  How  many  40-watt  electric  glow  lamps  can  be  run  on  a 
110- volt  circuit  with  an  expenditure  of  48  amperes  of  current  ? 

Brown  and  Sharpe  Wire  Table 

The  unit  chosen  for  this  table  is  a  copper  wire  .1  inch  in 
diameter.  This  is  called  No.  10  wire  and  has  the  following 
characteristics:  No.  10,  B  &  S  wire,  diameter,  .1  inch;  area  in 
circular   mils,  10,000  cm.;  resistance,  per   1000   ft.,  1  ohm; 


ELECTRICAL  WORK  249 

weight  per  1000  ft.,  31. o  lb.  Since  this  table  was  made,  the 
standard  has  changed  slightly,  so  that  at  present  No.  10  wire 
is  .1019  inch  in  diameter  and  the  other  vahies  are  changed 
l)roportionately,  but  for  all  ooiumercial  work  the  values  as 
originally  given  are  sufficiently  accurate. 

The  table  is  so  arranged  that  the  area  of  the  wire  is  doubled 
every  three  gauges  down  and  halved  every  three  numbers  up. 

Example.  —  Find  the  area  of  No.  7  wire. 

No.  7  is  three  numbers  below  No.  10,  whose  area  is  10,000  cm.,  so  that 
its  area  is  2  x  10,000  =  20,000  cm. 

Area  of  No.  4  wire  =  2  x  20,000  =  40,000  cm. 
Area  of  No.  13  wire  =  I  x  10,000  =  5000  cm. 

The  resistance  is  reduced  to  one  half  every  three  numbers 
down  and  doubled  every  three  numbers  up.  The  weight 
doubles  every  three  numbers  down  and  halves  every  three 
numbers  up. 

Example.  — 

R  per  1000  ft.  of  No.  10  H  &  S  wire  equals  1  ohm 
R  per  1000  ft.  of  No.    7  H  &  S  wire  equals  .5  ohm 
R  per  1000  ft.  of  No.    4  H  &  S  wire  equals  .25  ohm 
R  per  1000  ft.  of  No.  13  B  &  S  wire  equals  2  ohms 
R  per  1000  ft.  of  No.  16  B  &  S  wire  equals  4  ohms 

Example.  — 

TTper  1000  ft.  of  No.  10  B  &  S  wire  equals  31.5  lb. 
W  per  1000  ft.  of  No.    7  B  &  S  wire  equals  (k^  lb. 
IF  per  1000  ft.  of  No.    4  B  &  S  wire  equals  126  lb. 
TKper  1000  ft.  of  No.  13  B  &  S  wire  equals  15.8  lb. 
IK  per  1000  ft.  of  No.  16  B  &  S  wire  equals  7.9  lb. 

If  the  gauge  number  is  not  three  or  a  multiple  of  three  be- 
low or  above  No.  10,  get  the  area,  resistance,  or  weight  desired 
which  is  less  than  the  value  for  the  wire  required,  and  if  it  is 
one  below  the  required  number,  multiply  by  1.26,  and  if  two 
below,  by  1.59. 


250  VOCATIONAL  MATHEMATICS 

In  computing  the  resistance  and  weight  of  cables  the  follow- 
ing formula  is  used : 

Resistance  =  R  = in  ohms  per  1000  ft. 

cm. 

Weight  =  .00305  x  cm.  in  lb.  per  1000  ft. 

EXAMPLES 
Find  the  resistance,  area,  and  weight  of  the  following  wires  : 

1.  No.  16.  6.   No.  14. 

2.  No.  13.  7.   No.  15. 

3.  No.  7.  8.   No.  10. 

4.  No.  3.  >  9.   No.  2. 

5.  No.  1.  10.   No.  11. 


PART  IX  — MATHEMATICS   FOR   MACHINISTS 

CHAPTER  XVI 
MATERIALS 

Every  machinist  should  be  familiar  with  the  strength  and 
other  properties  of  the  materials  that  he  uses  —  such  metals  as 
cast  iron,  wrought  iron,  steel,  copper,  bronze,  and  brass. 

Cast  Iron.  —  Since  much  of  the  machinist's  work  is  on  cast  iron,  he 
should  know  something  of  its  nature  and  manufacture.  Iron  ore  as 
found  in  the  earth  generally  contains  many  impurities,  such  as  silicon, 
sulphur,  phosphorus,  manganese,  combined  carbon,  and  graphitic  carbon. 
To  free  the  iron  from  the  grosser  impurities,  the  ore  is  crushed  and  mixed 
with  coke  and  limestone  and  intense  heat  applied  in  a  blast  furnace.  The 
melted  iron,  being  heavier  than  the  other  materials,  falls  to  the  bottom  of 
the  furnace.  When  a  sufficient  quantity  has  accumulated,  it  is  allowed 
to  flow  out  of  a  tap  hole  into  molds  of  sand.  After  it  has  cooled  it  is 
broken  into  lengths  suitable  to  be  remelted  in  foundries  and  made  into 
iron  castings.     Such  iron  is  called  pig  iron. 

During  the  process  of  smelting  in  the  blast  furnace,  the  liquid  iron 
combines  with  a  considerable  quantity  of  carbon,  sulphur,  silicon,  phos- 
phorus, and  manganese  from  the  impurities  in  the  ore  and  coke.  Some 
of  the  carbon  combines  with  the  iron  chemically  and  forms  iron  carbide, 
while  the  remainder  exists  in  the  iron  as  a  mixture  of  carbon  and  is 
known  as  graphite.  The  amount  of  carbon  may  weaken  the  iron  by 
making  it  soft,  and  it  may  also  make  the  iron  too  brittle  to  work.  So 
the  man  in  charge  of  the  foundry  must  use  his  judgment  in  mixing 
different  grades  and  quantities  of  pig  iron  to  obtain  a  casting  of  the  de- 
sired strength,  hardness,  toughness,  and  clearness  of  grain. 

Castings.  —  Machines  are  made  of  iron  castings,  forgings,  steel  parts, 
etc.  Castings  and  forgings  can  be  distinguished  from  each  other  by  the 
appearance  of  the  fractures  in  them.  After  the  machines  are  designed 
and  the  wooden  patterns  made  in  the  pattern  shop,  the  patterns  are  sent  to 

251 


252  VOCATIONAL  MATHEMATICS 

the  foundry,  where  aa  impression  of  the  machine  is  made  in  sand.  Dur- 
ing this  operation  of  molding,  the  sand  is  confined  in  an  iron  or  wooden 
device  of  two  or  more  parts  called  a  flask.  The  lower  or  bottom  part  of 
the  flask  is  the  drag  or  nowel,  while  the  top  or  upper  part  is  the  cope; 
and  other  parts  are  the  checks. 

Sometimes  pig  iron  and  old  scrap  iron  are  melted  together  in  a  furnace 
called  a  cupola.  The  liquid  iron  is  taken  from  the  furnace  in  ladles  and 
poured  into  different  molds.  As  the  hot  iron  flows  into  the  mold  and 
cools,  it  becomes  solid  and  takes  the  shape  of  the  mold. 

When  the  castings  are  removed  from  the  mold  they  present  a  rough 
surface  and  have  to  be  cleaned  and  smoothed  or  "machined"  before 
they  can  be  put  to  the  use  intended  for  them.  They  are  cleaned  in 
various  ways  —  by  means  of  emery  wheels  and  revolving  wire  brushes, 
by  being  rotated  in  "  tumblers,"  by  chipping  with  pneumatic  chisels,  or 
by  means  of  a  sand  blast.  The  scales  on  the  castings  are  removed  by 
wetting  them  with  diluted  sulphuric  acid.  This  process  is  called  pickling. 
After  this  the  casting  is  attached  to  the  plate  or  table  of  the  machine 
tool  that  is  to  perform  the  necessary  work  upon  it.  Special  devices  are 
made  to  hold  castings  when  they  are  being  machined. 

Wrought  Iron.  —  One  of  the  valuable  qualities  of  wrought  iron  is  the 
comparative  ease  with  which  it  can  be  united  with  another  piece  by  weld- 
ing. When  two  pieces  of  wrought  iron  are  heated  to  a  white  heat,  they 
assume  a  viscous  condition,  and  when  hammered  together  become  united. 
Wrought  iron  differs  from  cast  iron  in  that  it  is  capable  of  assuming  any 
shape  under  the  hammer.  It  is  readily  made  from  cast  iron  by  heating 
in  &  puddling  furnace.  In  this  furnace  the  cast  iron  is  subjected  to  great 
heat  and  constant  stirring,  which  allows  the  carbon  to  pass  off  as  a  gas 
and  the  other  impurities  to  rise  to  the  surface,  where  they  can  be  removed. 
When  the  impurities  'are  removed  the  iron  is  hammered  to  remove 
particles  of  slag  and  then  rolled  in  order  to  make  it  more  compact.  After 
this  it  is  heated  again  and  rolled  into  bars  for  different  purposes. 

Wrought  iron  is  sometimes  case-hardened  when  it  is  used  in  machine 
parts  that  need  to  be  harder  than  the  common  iron.  After  the  piece  has 
been  finished  and  properly  sized  it  is  heated  a  bright  red  and  the  surface 
rubbed  with  prussiate  of  potash.  When  it  has  cooled  to  a  dull  red,  it  is 
immersed  in  water.  Three  parts  of  prussiate  of  potash  and  one  of  sal 
ammoniac  is  a  good  case-hardening  mixture.  The  temperature  (Fahren- 
heit) for  cherry  red  is  1832°.  If  there  are  holes  in  such  iron  work, 
the  hardening  by  this  process  reduces  them  slightly. 

Steel.  —  Steel  is  a  form  of  iron  which  contains,  as  a  rule,  more  car- 
bon and    other  elements  than  wrought  iron   and  less   than  cast  iron. 


MACHINISTS'   WORK  253 

There  are  many  grades  of  steel,  and  each  one  is  made  by  a  8i)ecial  process, 
steel  may  be  recognized  by  the  appearance  of  a  dark  spot  when  nitric 
acid  is  placed  on  its  surface.  The  darker  the  spot,  the  harder  the  steel. 
Iron,  on  the  contrary,  shows  no  sign  when  touched  with  nitric  acid, 
(iootl  steel  will  not  stand  a  high  heat,  but  will  crumble  under  the  hammer 
blow  at  a  bright  red  heat,  while  at  a  moderate  heat —  a  full  dull  red  or 
cherry  red  —  it  may  be  drawn  out  to  a  fine  edge  tool.  Steel  that  has 
once  been  overheated  or  burned  cannot  be  restored. 

Steel  for  cutting-tools  on  lathes  and  i)laners  should  be  drawn  to  a  straw 
color,  or  430^,  while  for  wood  tools,  taps,  and  dies,  dark  straw  color,  or 
41(f ;  for  chisels  for  chipping,  brown  yellow  or  SW ;  for  springs,  dark 
purple,  or  650°. 

Steel  may  be  softened  or  annealed  if  heated  to  a  low  red  and  placed  in 
a  box  of  slaked  lime  and  well  covered,  or  in  a  box  of  fine  bonedust,  care 
being  taken  in  either  case  to  cover  the  piece  all  around  and  on  top  to  a 
depth  of  not  less  than  one  and  one  half  inches. 

Copper.  —  Copper  is  used  to  a  great  extent  because  it  may  be  easily 
forged  when  cold.  It  may  also  be  pressed  into  different  shapes  by  means 
of  molds.  Its  strength  is  greatly  increased  by  hammering  and  rolling. 
It  is  used  principally  in  wires  and  plates. 

Brass.  —  Brass  is  an  alloy  or  mixture  of  copper  and  zinc.  Its  tensile 
strength  is  nearly  equal  to  that  of  copper. 

Weight  of  Bars  of  Steel.  —  The  weight  of  bars  of  steel,  as 
they  are  usually  made,  is  found  by  multiplying  the  area  of 
the  ci'oss  section  or  end  in  inches  by  the  length  in  inches,  and 
multiplying  the  resulting  nural^er  of  cubic  inches  in  the  bar 
by  0.3.  This  will  give  practically  an  accurate  result,  since  all 
bars,  unless  otherwise  ordered,  will  be  rolled  or  hammered 
slightly  "full"  to  the  dimensions  given,  or  a  bar  that  is 
"  full "  to  size  may  be  trimmed  to  exact  dimensions  if  neces- 
sary. Whereas,  if  the  bar  is  slightly  under  size,  it  cannot 
easily  be  made  larger. 

To  find  the  weight  of  a  triangular  bar  of  steel,  multiply  the 
area  of  the  base  in  square  inches  by  the  height  in  inches  and 
then  by  0.3.  The  base  of  the  triangular  bar  is  found  by  multi- 
plying the  length  of  one  of  the  bars  or  sections  by  one  half  the 
l)erpendicular  height;  that  is,  by  the  distance  to  the  opposite 
vertex  of  the  cross  section. 


254  VOCATIONAL  MATHEMATICS 

EXAMPLES 

1.  What  is  the  weight  of  a  triangular  bar  of  steel  when  the 
base  contains  16  sq.  in.  and  the  height  is  6  ft.  ? 

2.  What  is  the  weight  of  a  triangular  bar  of  steel  when  the 
base  contains  13  sq.  in.  and  the  height  is  4  ft.  ? 

3.  What  is  the  weight  of  a  triangular  bar  of  steel  when  the 
base  contains  21  sq.  in.  and  the  height  is  lo  feet  ? 

4.  What  is  the  weight  of  a  triangular  bar  of  steel  when  the 
base  contains  23  sq.  inches  and  the  height  is  17  feet  ? 

5.  What  is  the  area  of  the  base  or  section  of  a  triangular 
bar  one  side  of  which  is  8  inches  and  the  altitude  6  inches  ? 

6.  What  is  the  area  of  the  base  or  section  of  a  triangular 
bar  one  side  of  which  is  11  in.  and  the  altitude  9  in.  ? 

7.  What  is  the  weight  of  a  triangular  bar  of  steel  one  side 
of  which  and  altitude  of  whose  base  or  section  are  respectively 
13  in.  and  11  in.,  and  the  length  of  the  bar  23  ft.  ? 

8.  What  is  the  weight  of  a  triangular  bar  whose  section  is 
24  sq.  in.  and  whose  length  is  22  ft.  ? 

9.  The  weight  of  a  cast-iron  wheel  is  approximately  sixteen 
times  as  heavy  as  the  white  pine  pattern  from  which  it  is  cast. 
What  is  the  probable  weight  of  a  casting  if  the  pattern  for  it 
weighs  2J  pounds  ? 

10.  A  white  pine  pattern  weighs  12.5  pounds.  What  will  be 
the  weight  of  an  iron  casting  from  it  ?     (Use  data  in  Ex.  9.) 

11.'  If  the  weight  of  a  brass  casting  is  approximately  fifteen 
and  a  half  times  that  of  its  white  pine  pattern,  what  will  be 
the  weight  of  a  casting  if  the  pattern  weighs  15  oz.  ? 

12.  A  white  pine  pattern  weighs  1.75  pounds.  What  will  be 
the  weight  of  50  brass  castings  made  from  it  ?  (Use  data  in 
Ex.  11.) 

13.  Since  the  shrinkage  of  brass  castings  is  about  ^  inch  in 
10  inches,  what  length  would  you  make  the  pattern  for  a  brass 
collar  which  is  required  to  be  6  inches  long  ? 


CHAPTER   XVII 

LATHES 

The  Engine  Lathe.  —  The  lathe  is  one  of  the  most  important 
machines  in  the  machine  shop.  There  are  different  kinds  of 
lathes,  each  adapted  to  certain  kinds  of  work,  the  engine 
lathe  being  the  most  important.  The  motion  of  the  tool  is 
controlled  by  power  speed;  that  is,  the  tool  is  moved  automati- 
cally parallel  with  and  at  right  angles  to  the  center  line  of  the 
lathe  spindle.  Most  lathes  are  furnished  with  a  series  of 
clutch  gears  and  lead  screws  by  means  of  which  threads  of 
different  pitch  may  be  cut.  All  lathes  have  a  series  of  stepped 
cones  in  order  to  obtain  a  variety  of  speeds  which  are  neces- 
sary in  order  to  work  on  hard  and  soft  metals  and  to  obtain 
constant  surface  speeds  for  different  diameters.  The  slower 
speed  makes  the  deeper  cut. 

Size.  —  The  size  of  the  lathe  is  expressed  by  stating  the 
length  of  the  bed  and  the  largest  diameter  it  will  swing  on 
centers.  The  swing  is  found  by  measuring  from  the  point  of 
the  headstock  center  to  the  ways  on  the  bed  and  then  multi- 
plying by  2.  The  English  measurement  is  from  the  center  to 
the  way.  The  next  important  measurement  is  the  length  of 
the  bed,  which  is  the  entire  amount  of  distance  the  tailstock 
will  move  backward.  If,  however,  accuracy  is  desired  in  this 
measurement,  the  figure  given  should  be  the  distance  between 
the  two  centers  when  the  tailstock  is  in  its  extreme  backward 
position,  as  the  lathe  will  turn  no  longer  piece  than  will  go 
between  centers. 

Gear  and  Pitch.  —  Lathes  that  will  cut  a  thread  the  same 
pitch  as  the  lead  screw,  with  gears  having  the  same  number 
of  teeth  on  the  stud  and  screw,  are  called  geared  even.     If  a 

255 


256 


VOCATIONAL  MATHEMATICS 


lathe  will  not  do  this,  find  what  thread  will  be  cut  with  even 
gears  on  both  stud  and  lead  screw  and  consider  that  as  the 
pitch  of  the  lead  screw. 

Adjusting  Gears.  —  A   simple  or  single-geared   lathe  is    one 
having  a  straight  train  of  gearing  from  its  spindle  to  its  feed 


Engine  Lathe 


screw,  excepting  intermediate  gears,  which  only  serve  as  idlers 
to  take  up  the  distance  between  the  driver  and  driven  gears 
or  spindle  and  screw  gears.  Index  plates  are  usually  found 
on  lathes  giving  the  change  gear  used  for  different  threads, 


LATHES  257 

but  when  threads  are  called  for  that  are  not  indexed,  or  when 
those  ending  in  fractions  are  to  be  cut,  the  machinist  must 
make  his  own  figures. 

Refer  to  the  screw  cutting  table  and  see  what  number  of 
threads  to  an  inch  are  cut  with  equal  gears.  This  number  is 
the  number  of  turns  to  an  inch  that  we  a^ume  the  lead  screw 
has,  no  matter  what  its  real  number  of  turns  to  an  inch  is. 
Write  above  the  line  the  number  of  turns  to  an  inch  of  the 
lead  screw  and  below  the  line  the  number  of  turns  to  an  inch 
of  the  screw  to  be  threaded,  thus  expressing  the  ratio  in  the 
form  of  a  fraction,  the  lead  screw  being  the  numerator  and  the 
screw  to  be  threaded  the  denominator.  Now  find  an  ecjual 
fraction  in  terms  that  represent  numbers  of  teeth  in  available 
gears.  The  numerator  of  this  new  fraction  will  be  the  spindle 
or  stud  gear  and  the  denominator  the  lead  screw  gear.  The 
new  fraction  is  usually  found  by  multiplying  the  numerator 
and  denominator  of  the  first  fraction  by  the  same  number. 

PlxAMPLE.  —  It  is  required  to  cut  a  screw  having  11^  threads 
l)er  inch. 

The  index  gives  48  to  48  cuts  4  threads  per  inch. 

_L  X  -*  =  — 
Hi      6       69 

Put  the  24-tooth  gear  on  the  stud  and  the  tt9-tooth  gear  on  the  lead 
screw  to  cut  11^  threads  per  inch. 

EXAMPLES 

1.  What  gears  should  be  used  to  cut  a  screw  having  16 
threads  per  inch,  if  a  40-gear  on  the  stud  and  an  80-gear  on 
the  screw  will  cut  8  threads  to  the  inch  ? 

2.  What  geai's  should  be  used  to  cut  a  screw  having  3 
threads  per  inch,  if  a  48-gear  on  the  stud  and  a  56-gear  on  the 
screw  will  cut  14  threads  to  the  inch  ? 

*  If  multiplying  by  6  will  not  give  the  gears  available,  use  any  other 
number. 


258  VOCATIONAL  MATHEMATICS 

3.  What  gears  should  be  used  to  cut  a  screw  32  threads  per 
inch,  if  the  pitch  of  the  lead  screw  is  12  ? 

4.  What  gears  should  be  used  to  cut  the  following  threads 
per  inch,  if  the  pitch  of  the  lead  screw  is  12  ? 

a.  36  thread*  d.   64  threads 

b.  42  threads  e.   3\  threads 

c.  56  threads  /   3^  threads 

g.   12  threads 

Compound  Lathes.  —  The  term  compound  applied  to  a  lathe 
means  that  in  its  train  of  gearing  from  its  spindle  to  lead 
screws  there  is  a  stud  or  spindle  having  two  different  sized 
gears,  both  connected  in  such  a  way  as  to  change  the  link  of 
revolution  between  the  spindle  and  the  lead  screw  to  a  different 
number  of  revolutions  from  that  which  would  take  place  if 
the  straight  line  of  gears  were  used. 

First  Method. — Write  the  number  of  turns  to  an  inch  of  the 
lead  screw  as  the  numerator  of  a  fraction  and  the  turns  of  the 
screw  to  be  threaded  as  the  denominator.  Factor  this  fraction 
into  an  equal  compound  fraction.  Change  the  terms  of  this 
compound  fraction  either  by  multiplying  or  dividing  into 
another  equal  compound  fraction  whose  terms  represent  num- 
bers of  teeth  in  available  gears.  Then  the  two  terms  in  the 
numerator  represent  the  number  of  teeth  in  the  gears  to  be 
used  as  drivers  and  those  in  the  denominator  the  gears  to  be 
used  as  driven  gears. 

Example.  • —  It  is  required  to  cut  a  screw  having  3^  inches 
lead  or  -^^  turns  to  an  inch.  The  lead  screw  is  1^  inches  lead 
or  I  turns  to  an  inch. 

3  '  13      3x4 
Multiply  numerator  and  denominator  by  5, 


2  X  (13  X  5)  _  2  X  65 
1  X  (12  X  6)      1  X  60 

>r  ,..  1  *  a' A  •     .      V.    o.  (24  X  2)  X  65     48  X  65 

Multiply  numerator  and  denominator  by  24,  (24  x  1)  x  60  "  24x60 


LATHES  259 

The  48-tootli  and  the  05-tooth  gears  will  be  tlie  drivers  and  the  24- 
tooth  and  GO-tooth  gears  the  driven. 

NoTK.  —  Any  multiplier  may  be  used  to  obtain  the  gear  that  is  avail- 
able. 

Second  Method.  —  Another  satisfactory  method  of  working 
out  the  change  gears  is  by  proportion.  If  it  is  desirable  to 
cut  a  screw  having  10  threads  per  inch  and  the  lead  screw 
has  6  threads  per  inch,  the  first  two  terms  of  the  proportion 
would  be  10:  6.  As  a  rule,  the  smallest  gear  in  the  gear  box 
is  used  on  the  spindle,  if  it  will  serve  the  purpose,  and  as  the 
number  of  teeth  on  this  gear  is  generally  a  multiple  of  the 
number  of  threads  per  inch  on  the  lead  screw,  in  the  present 
case  it  would  probably  be  24.  As  the  number  of  teeth  on  the 
lead  screw  is  to  be  the  third  term  of  the  proportion,  and  as 
this  is  unknown,  x  is  used  to  represent  it,  and  then  the  pro- 
portion is  10:  6:  :  x:  24.  By  multiplying  the  first  and  fourth 
terms  together  and  the  second  and  third  terms  together,  the 
result  is  6x  =  240.  Then  ic  =  40,  the  number  of  teeth  on  the 
screw  gear. 

If  the  lathe  is  compound  geared,  it  is  necessary  to  find  the 
proportional  speed  of  spindle  and  stud.  If  the  stud  makes 
three  quarters  of  a  revolution  while  the  spindle  makes  a  com- 
plete revolution,  it  is  necessary  to  use  a  gear  on  the  screw 
with  but  three  quarters  the  number  of  teeth  represented  by  x 
in  the  proportion. 

Example.  —  What  gear  should  be  used  on  the  screw  of  a 
compound  geared  lathe  with  the  stud  turning  only  three 
quarters  as  fast  as  the  spindle,  in  order  to  cut  a  screw  having 
13  threads  per  inch,  if  the  lead  screw  has  6  threads  per  inch 
and  the  stud  gear  48  teeth  ? 

13  :  6  :  :  a: :  48 
8 
Cancelling,  13  :  ^  :  :  x  :  ^  =  104 
Three  quarters  of  104  =  78 

With  a  78-tooth  gear  on  the  screw,  a  48-tooth  gear  on  the  stud  of  the 
compound  geared  lathe  will  cut  a  screw  having  13  threads  per  inch. 


260  VOCATIONAL  MATHEMATICS 

Note.  —  If  the  stud  turned  but  one  half  as  fast  as  the  spindle,  then  a 
gear  should  be  used  on  the  screw  with  one  half  as  many  teeth  as  shown 
under  the  method  for  simple  geared  lathes.  , 

QUESTIONS   AND  EXAMPLES 

1.  Is  the  lathe  in  the  classroom  simple  or  compound 
geared  ? 

2.  What  gears  should  be  used  to  cut  a  screw  having  18 
threads  per  inch,  if  a  40-gear  on  the  stud  and  an  80-gear  on 
the  screw  will  cut  8  threads  to  the  inch  ? 

3.  How  many  threads  per  inch  has  the  lead  screw  ?  Is  it 
a  square,  V,  or  acme  thread  ?     Is  it  right  or  left  hand  ? 

4.  What  gears  should  be  used  to  cut  a  screw  having  6 
threads  per  inch,  if  a  48-gear  on  the  stud  and  a  56-gear  on  the 
screw  will  cut  14  threads  to  the  inch  ? 

5.  How  many  gears  are  there  between  the  spindle  gear  and 
the  lead  screw  gear  ? 

6.  If  the  pitch  of  the  lead  screw  is  16,  what  gears  should 
be  used  to  cut  a  screw  with  38  threads  per  inch  ? 

7.  Has  this  lathe  reversing  gears?  If  not,  state  briefly 
how  the  reversing  is  accomplished. 

8.  If  the  pitch  of  the  lead  screw  is  18,  what  gears  should 
be  used  to  cut  a  screw  with  44  threads  per  inch  ? 

9.  Put  even  gears  on  the  first  change  gear  stud  and  lead 
screw  and  put  a  smooth  round  piece  of  wood  on  the  lathe 
centers.  Clamp  a  pencil  on  the  tool  post  so  that  it  will  mark 
on  the  wood,  then  turn  the  lathe  spindle  until  the  carriage  has 
moved  1  inch.  How  many  threads  did  the  pencil  draw  on  the 
wood  ? 

10.  What  gears  should  be  used  to  cut  the  following  threads 
per  inch,  if  the  pitch  of  the  lead  screw  is  18  ? 

a.  42  threads  c.  16  threads 

h.  8  threads  d.  5^  threads 


LATHES  261 

11.  What  is  the  true  pitch  of  the  lead  screw  ?  Is  it  the 
same  as  the  actual  pitch  ? 

12.  A  1"  bolt  is  to  have  8  threads  per  inch  cut  in  it.  If  a 
56-tooth  gear  is  on  the  lead  screw,  what  gear  must  be  put  on 
the  stud  ? 

13.  A  lathe  has  a  feed  rod  turning  at  the  same  rate  as  the 
lead  screw,  while  the  carriage  travels  one  quarter  as  fast  as 
it  would  when  screw  cutting.  If  geared  for  12  threads  per  inch 
and  the  feed  shaft  is  used,  what  will  be  the  feed  in  fractions  of 
an  inch  per  revolution  ? 

14.  At  64  revolutions  per  minute  (R.  P.  M.)  how  long  will  it 
take  to  make  a  roughing  cut  with  y\"  feed  and  a  finishing  cut 
with  y\"  feed,  if  both  cuts  are  21"  long  and  1  minute  is 
allowed  for  changing  tools  ? 

To  Cut  Double  or  Multiple  Threads 

In  modern  machine  construction  there  are  many  studs, 
screws,  and  feed  rods  having  threads  for  rapid  travel,  and 
instead  of  a  single  spiral  thread,  there  are  two  and  sometimes 
three  spirals.  If  a  machinist  is  called  upon  to  replace  or 
duplicate  such  a  thread,  the  method  would  not  be  to  cut  the 
multiple  threads  at  one  time  but  to  cut  one  thread  at  a  time. 
If  the  pitch  of  the  double  thread  is  measured,  the  pitch  of 
every  second  thread  will  be  measured  and  the  lathe  set  for  4 
threads  per  inch.  The  cut  of  4  threads  is  chased  out  to  size 
and  the  lathe  left  geared  and  the  tool  unchanged,  and  by  turn- 
ing the  gears  on  the  lathe  spindle  one  half  revolution,  the  tool 
position  for  beginning  the  second  thread  is  gained. 

Before  fixing  the  gear  wheel  position,  the  carriage  should  be 
reversed  in  the  position  of  the  cut,  thus  taking  up  all  lost 
motion  in  screws,  lathe  nuts,  and  gears.  Then  make  the  exact 
position  of  the  mesh  on  the  teeth  of  the  spindle  and  on  the 
lead  screw  and  count  the  number  of  teeth  on  the  spindle  gear, 
which   equals   one  half  the  entire   number,   and   make   this 


262  VOCATIONAL  MATHEMATICS 

tooth.  Now  take  off  the  spindle  and  turn  the  lathe  one  half  a 
revolution,  bringing  the  second  marked  tooth  to  the  position 
of  the  first,  and  the  lathe  is  then  ready  for  cutting  the  second 
thread.  It  is  necessary  in  cutting  multiple  threads  to  select  a 
driving  gear  wheel  having  a  number  of  teeth  exactly  divisible 
by  the  number  of  threads  cut. 


Machine  Speeds 

Eveiy  casting  is  cut  into  a  definite  shape  by  removing  a  certain  portion 
of  it,  called  a  cut  or  turning,  by  one  of  the  machine  shop  tools.  The 
casting  must  be  cut  in  the  most  economical  way  and  in  the  shortest  pos- 
sible time.  The  tool  that  does  the  cutting  must  attain  the  highest  pos- 
sible speed,  but  there  is  a  limit  to  the  speed  of  a  tool  on  account  of  the 
heat  generated  as  it  moves  against  the  casting.  If  too  great  speed  is 
used,  the  heat  generated  takes  the  temper  out  of  the  tool,  renders  it  use- 
less, and  causes  the  casting  to  expand.  The  effect  of  the  expansion  on 
the  casting  is  to  make  it  no  longer  true.  The  machine  that  is  doing  the 
cutting  should  be  accurate  to  .005  of  an  inch. 

The  cutting  capacity  of  a  machine  depends  on  (1)  the  speed 
of  the  cut,  (2)  the  distance  traversed  by  the  tool  in  passing 
from  one  cutting  portion  to  the  next,  (3)  the  depth  of  cut,  i.e., 
the  thickness  of  the  strip  removed  from  the  casting. 

The  volume  of  metal  removed  from  a  casting  may  be  calcu- 
lated as  follows: 

The  volume  of  metal  removed  from  good  steel  in  one  minute 
equals  :  Cutting  speed  (length)  times  the  feed  (width)  times 
the  depth  of  cut  (thickness). 

Example.  —  If  the  speed  of  a  cut  is  18  ft.  per  minute,  the 
feed  .06  in.,  and  the  depth  of  cut  is  \  in.,  what  is  the  volume 
of  metal  removed  in  one  hour  ? 

18'  X  12"  X  .06  X  .25  =  3.24  cu.  in. 
weight  of  1  cu  in.  of  good  steel  =  .277  lb. 
3.24  X  .277  =  0.897  lb. 
.897  X  60  =  53.8  lb.  in  1  hr.     Ans. 


LATHES  263 

Speeds  for  Different  Metals.  —  Various  materials,  such  as  iron,  copper, 
ami  wrought  iron,  possess  different  standards  of  hardness.  A  hanler 
metal  will  wear  away  the  tool  and  strain  the  machine  more  than  a  softer 
metal.  Therefore,  there  are  different  sj^eeds  for  each  metal.  With  car- 
bon steel  cutting  tools,  the  surface  speed  varies  from  30  to  40  feet  per 
minute  for  cast  iron,  wrouglit  iron,  and  soft  steel ;  15  to  26  feet  for  well 
annealed  tool  steel,  and  from  60  to  80  feet  per  minute  for  brass,  while 
the  speed  for  dies  and  taps  varies  from  12  to  18  feet  per  minute  on  steel, 
and  from  30  to  50  feet  per  minute  on  brass,  depending  upon  the  quality 
of  the  metal  and  the  shape  of  the  piece.  With  cutting  tools  of  high-speed 
steel  these  speeds,  except  for  dies  and  taps,  can  be  nearly  doubled. 

Cast  iron  should  be  worked  at  a  speed  which  is  j^j  to  j'j  of  that  for 
copper,  or  J  to  1*5  of  that  for  wrought  iron. 

Net  Power  for  Cutting  Iron  or  Steel 

To  find  the  net  power  for  cutting  cast  iron  and  steel,  mul- 
tiply the  section  of  cut  or  chip  in  square  inches  by  230,000 
pounds  for  steel,  or  168,000  pounds  for  cast  iron,  to  get  the 
pressure  on  the  tool ;  and  multiply  this  product  by  the  cutting 
speed  in  feet  per  minute,  and  divide  the  result  by  33,000  to 
obtain  the  horse  power  required. 

Example.  —  Steel  is  being  cut  with  \"  cut,  1-64"  feed  at 
a  speed  of  20'  per  minute.     What  is  the  H.  P.  ? 

\"  X  1-64"  =  .0039  sq.  in. 

.0039  X  230,000  lb.  =  897  lb.,  pressure  on  tool 

897  X  20  =--  17,940 

17,940  -T-  33,000  =  .64  horse  power.     Ans. 

Example.  —  If  the  cutting  speed  at  the  rim  of  iron  stock 
should  be  40  feet  per  minute,  at  what  speed  should  the  lathe 
(spindle)  be  driven  for  a  piece  of  stock  3"  in  diameter  ? 

3"  diameter 

8  X  8.1416  =  9.4248",  nearly  9.6",  or  f ' 

40'  H- 1' 

40  X  J  =  ifft  =  53  R.  P.  M.     Ans. 


264  VOCATIONAL  MATHEMATICS 


EXAMPLES 

1.  What  is  the  amount  of  metal  removed  in  one  hour  from 
a  casting  with  a  cutting  speed  of  69'  per  minute  and  a  feed 
of  j\"  and  a  depth  of  -f  ? 

2.  As  a  tool  will  stand  a  cutting  speed  of  35  feet  per 
minute  when  turning  cast  iron,  how  many  revolutions  per 
minute  should  the  lathe  spindle  make  when  a  piece  of  cast 
iron  8''  in  diameter  is  being  turned  ? 

3.  Write  the  formula  for  the  feed  of  a  lathe  tool,  making 
JV"=  number  of  revolutions,  D  =  distance  the  tool  moves,  and 
i^=the  feed. 

4.  A  lathe  tool  moves  2.1-"  along  the  work  in  one  minute, 
and  the  speed  of  the  lathe  is  400  R.  P.  M.  What  is  the  feed  ? 
(Use  the  formula.) 

5.  In  turning  up  in  the  lathe  a  gun  metal  valve,  4  inches 
in  diameter,  it  is  desirable  that  the  surface  speed  shall  not 
exceed  45  ft.  per  minute.  How  many  revolutions  per  minute 
may  the  w^heel  make  ? 

6.  Find  the  time  required  and  the  speed  of  the  lathe  in 
turning  one  20-foot  length  of  3-inch  wrought  iron  shafting, 
one  cut,  traverse  28  per  inch,  cutting  speed  20  ft.  per  minute, 
no  allowance  being  made  for  grinding  or  breaking  of  tools  or 
for  setting  stays. 

7.  How  many  revolutions  per  minute  may  be  made  in 
turning  np  a  steel  shaft  6  inches  in  diameter,  if  the  surface 
speed  must  not  exceed  12  ft.  per  minute  ? 

8.  A  piece  of  tool  steel  1|''  in  diameter  is  turned  in  a 
lathe  at  74  K.  P.  M.     What  is  the  cutting  speed  ? 

9.  What  is  the  amount  of  metal  removed  in  one  hour  from 
a  casting  with  a  cutting  speed  of  64  ft.  per  minute  and  a  feed 
of  jY  and  a  depth  of  i"  ? 

10.   If  a  tool  will  stand  a  cutting  speed  of  237  F.  P.  M.  when 


LATHES  265 

turning  cast  iron,  how  many  R.  1*.  M.  should  the  hithe  (spindle) 
make  when  a  piece  of  cast  iron  iV  in  diameter  is  being  turned  ? 

11.  A  valve  yoke,  stem  2"  in  diameter  is  being  turned  in  a 
lathe.  If  the  lathe  spindle  makes  50  H.P.  M.,  what  is  tlie 
cutting  speed  in  F.  P.  M.  ? 

12.  What  number  of  revolutions  must  a  lathe  spindle  make 
to  cut  15  ft.  per  minute,  in  turning  up  an  iron  shaft  7J  inches 
in  diameter  ? 

13.  (a)  With  a  tool  steel  that  can  stand  a  cutting  speed  of 
30'  per  minute,  how  many  revolutions  per  minute  may  a 
lathe  be  run  in  turning  one  cut  off  a  piece  of  shafting  15"  in 
diameter  ? 

(b)  To  hold  the  same  cutting  speed,  how  many  revolutions 
per  minute  would  be  required  if  the  shaft  were  but  7^"  in 
diameter  ? 

14.  In  turning  a  cast  iron  piston  head  15  inches  in  diameter 
in  the  lathe,  it  is  desired  that  the  surface  speed  shall  not  ex- 
ceed 15  ft.  per  minute.  How  many  revolutions  per  minute 
may  the  work  make  ? 

15.  How  many  revolutions  per  minute  should  the  lathe 
spindle  make  in  turning  up  a  cast  iron  pulley  33J  inches  in 
diameter,  at  a  cutting  speed  of  15  feet  per  minute  ? 

Table  of  Surface  Speeds 

The  table  on  page  266  has  been  computed  to  facilitate  the 
figuring  of  speeds  for  machines  and  is  used  as  follows : 

Having  first  determined  the  proper  surface  speed,  refer  in 
the  table  to  the  column  of  revolutions  per  minute  corresponding 
to  this  surface  speed,  and  in  this  column  opposite  the  diameter 
corresponding  to  that  of  the  work  under  consideration  will  be 
found  the  required  revolutions  per  minute  of  spindle  or  work. 
Other  surface  speeds  than  those  for  which  the  table  is  com- 
puted can  be  readily  obtained  from  the  table  by  multiplying 
or  dividing,  as  the  case  may  require. 


266 


VOCATIONAL   MATHEMATICS 


Table  of  Surface  Speeds 


32^ 


3TJ 


47^ 


DiAM. 

Bevolntions  per 

Mimitc 

■i^ 

3667.8 

3973.5 

4279.1 

4584.8 

4890.4 

5196.1 

6501.8 

6807.4 

6113.0 

l\ 

1838.9 

1986.7 

2139.5 

2292.4 

2445.2 

2598.0 

2750.9 

2903.7 

3056.5 

^ 

1222.6 

1324.5 

1426.3 

1528.2 

1630.1 

1732.0 

1833.9 

1936.8 

2037.7 

i 

916.9 

993.3 

1069.8 

1146.2 

1222.6 

1299.0 

1375.4 

1451.8 

1628.3 

A 

733.6 

794.7 

855.8 

916.9 

978.0 

1039.2 

1100.3 

1161.4 

1226.6 

t\ 

611.3 

662.2 

713.2 

764.1 

815.7 

866.0 

916.9 

967.9 

1018.8 

j\ 

523.9 

567.6 

611.3 

654.9 

698.6 

742.3 

785.9 

829.0 

873.3 

k 

458.4 

496.6 

534.9 

673.1 

611.3 

649.5 

687.7 

725.9 

764.1 

^ 

407.5 

441.5 

475.4 

509.4 

543.3 

677.3 

611.3 

645.2 

679.2 

t\ 

366.7 

397.3 

427.9 

458.4 

489.0 

519.6 

560.1 

680.7 

611.3 

M 

333.4 

361.2 

389.0 

416.9 

444.5 

472.3 

500.0 

627.9 

555.7 

f 

305.6 

331.1 

356.6 

382.0 

407.6 

433.0 

458.6 

483.9 

509.4 

il 

282.1 

305.6 

329.1 

352.3 

376.1 

399.7 

423.2 

446.7 

470.2 

tV 

261.9 

283.8 

305.6 

327.4 

349.3 

371.1 

392.9 

414.8 

436.6 

M 

244.5 

264.9 

285.1 

305.6 

326.0 

346.4 

366.7 

387.8 

407.5 

i 

229.2 

248.3 

267.4 

286.5 

305.6 

324.7 

343.8 

362.9 

382.0 

^ 

203.7 

220.7 

237.7 

254.7 

271.6 

288.6 

306.6 

322.6 

340.0 

f 

183.4 

198.6 

213.9 

229.2 

244.5 

259.8 

276.0 

290.3 

306.2 

ii 

166.7 

180.6 

194.5 

208.4 

222.3 

236.1 

250.0 

263.9 

277.8 

1 

152.8 

165.5 

178.3 

191.0 

203.7 

216.6 

229.2 

241.9 

254.2 

H 

141.0 

152.8 

164.5 

176.2 

188.0 

199.8 

211.6 

223.3 

234.8 

1 

130.9 

141.9 

152.8 

163.7 

174.6 

186.6 

196.4 

207.4 

218.2 

if 

122.2 

132.4 

142.6 

152.8 

163.0 

173.2 

183.4 

193.9 

203.8 

1 

114.6 

124.2 

133.7 

143.2 

152.8 

162.3 

171.9 

181.4 

191.0 

IrV 

108.2 

117.2 

126.2 

135.2 

143.2 

163.2 

162.2 

170.8 

180.4 

H 

101.9 

110.3 

118.8 

127.3 

136.8 

144.3 

152.8 

161.3 

170.0 

ii\ 

96.8 

104.8 

113.0 

121.0 

129.2 

137.0 

145.0 

152.8 

161.2 

H 

91.7 

99.3 

106.9 

114.6 

122.2 

129.9 

137.6 

145.1 

153.1 

lA 

87.5 

94.7 

102.1 

109.4 

116.8 

124.0 

131.2 

138.2 

146.0 

LATHES  267 


EXAMPLES 


1.  If  the  cutting  speed  of  soft  steel  is  25,  find  the  speed 
(spindle)  for  ly  stock. 

2.  If  the  cutting  speed  of  soft  steel  is  25,  find  the  speed 
(spindle)  for  f  J^"  stock. 

3.  If  the  cutting  speed  of  tool  steel  is  75,  find  the  speed 
(spindle)  for  ^4''  stock. 

4.  According  to  the  table  find  the  speed  of  (a)  soft  steel 
1^"  in  diameter;  (b)  soft  steel  ^"  in  diameter;  (c)  tool  steel 
2 J"  stock ;  (d)  tool  steel  2V'  stock ;  (e)  soft  steel  ^"  in 
diameter ;  (/)  soft  steel  if"  in  diameter. 


CHAPTER   XVIII 
PLANERS,   SHAPERS,   AND  DRILLING  MACHINES 

Planers.  —  The  planer  is  usually  one  of  the  heaviest  machines  in  the 
machine  shop.  Planing  is  rough  and  heavy  v^ork  and  stiffness  is  needed 
in  order  to  make  the  heavy  cuts  the  planer  usually  has  on  castings  and 
heavy  forgings.  In  many  shops  much  of  this  rough  surface  cutting  is 
saved,  however,  by  the  use  of  a  pickling  solution  of  one  part  of  sulphuric 
acid  to  eight  parts  of  water.  This  is  painted  over  the  casting  and,  after 
four  or  five  hours,  washed  off  with  clear  water,  removing  the  sand  and 
hard  grit  from  the  surface. 


Machine  Shop  Planer 

Care  must  be  taken,  during  the  operation  of  a  planer,  that  chips  and 
dirt  from  the  platen  are  not  swept  or  allowed  to  blow  into  the  V-ways,  as 
this  will  cause  injury  to  the  machine.  Great  care  must  be  taken,  also, 
that  oil  is  freely  supplied  to  the  machinery  under  the  planer  platen,  as 
these  parts  are  hidden  and  likely  to  be  neglected. 

A  planer  24  inches  by  24  inches  by  6  feet  will  consume  an  average  of 
0.085  horse  power  for  every  pound  of  cast  iron  removed  per  hour,  and  a 
consumption  of  0.066  horse  power  for  every  pound  of  n)achinery  steel 
removed  per  hour  under  good  operation  with  sharp  tools.     The  operator 

268 


PLANERS  AND   DRILLING   MACHINES  269 

of  a  planer  may,  by  poor  grinding  and  not  setting  his  tools,  waste  much 
power  in  driving  the  machine. 

The  cutting  speed  for  planers  is  about  the  same  as  for  lathe  tools,  the 
advantage  of  the  planer  being  ihat  heavier  cuts  can  be  produced  on 
account  of  the  rigid  support  of  the  platen. 

The  principal  objection  to  planing  machines  is  that  they  perform 
useful  work  in  only  one  half  of  the  motion.  When  the  work  is  drawn 
back,  no  cutting  takes  place.  Then  again,  siiice  the  forces  are  com- 
pletely changed  when  the  motion  is  changed,  the  slides  upon  which  the 
table  moves,  suffer.  In  this  way  accurate  work  is  difficult  on  account 
of  the  backlash.  The  return  on  capital  invested  in  such  machines  is 
smaller  than  in  the  case  of  machines  with  continuous  motion. 

Planer  and  Shaper.  —  The  planer  and  shaper,  and  such  modified  ma- 
chines as  the  slotting  machine  and  key  seater,  are  in  a  distinct  cla.ss.  Their 
use  is  to  machine  and  plane  irregular  surfaces  that  can  be  machined  by  a 
straight  line  cut.  The  cutting  tools  of  the  planer  and  shaper  are  practi- 
cally the  same  as  those  on  the  lathe.  In  the  planer  the  work  moves,  and 
vertical  and  lateral  feeds  are  given  to  the  tools.  In  the  sliaper  the  tools 
move  over  the  work,  lateral  to  the  work  and  vertical  to  the  tool.  The 
shaping  machine  is  designed  for  small  pieces  and  short  travels.  It  is 
nicely  adapted  for  cutting  grooves,  slots,  and  dovetails. 

Shaper  tools  should  be  kept  in  the  best  condition  as  to  shear,  clearance, 
and  cutting  edge,  as  they  are  called  upon  to  do  accurate  shaping  of  metal 
parts  of  machinery  not  possible  on  other  machines.  Shear  is  a  certain 
amount  of  angle  given  the  face  of  a  tool,  which  throws  its  cutting  edge 
forward  into  the  metal  to  be  cut.  Tools  without  clearance  drag  and  pull 
heavily  through  the  metal. 

For  work  on  the  shaper,  the  student  should  have  a  good  knowledge  of 
the  small  try-square  and  the  surface  block.  These  tools  are  constantly 
used,  the  square  showing  when  finished  pieces  are  square  with  the  shap- 
ing machine  vise  or  when  one  cut  is  square  with  another.  A  universal 
bevel  or  a  bevel  protractor  should  also  be  used,  enabling  angles  to  be  laid 
out  for  planing;  A  scriber  and  a  four-inch  outside  caliper  will  enable  the 
beginner  to  grasp  the  first  operations  of  shaping  and  planing.  The  ram 
should  be  adjusted  to  proper  length  of  stroke  to  save  time.  If  a  piece  of 
work  which  measures  two  and  three  fourths  inches  is  to  be  placed  on 
the  machine,  one  fourth  inch  is  sufficient  for  the  tool  to  enter  and  one 
eighth  inch  is  even  more  than  enough  to  allow  the  head  to  overreach. 
These  additions  then  give  us  a  total  stroke  of  three  and  one  eighth  inches, 
and  any  amount  over  this  loses  time  and  causes  unnece8.sary  wear  on  the 
machine. 


270  VOCATIONAL  MATHEMATICS 

EXAMPLES 

1.  If  a  planer  has  a  cutting  speed  of  30  ft.  per  minute  and 
a  return  speed  of  147  ft.  per  minute,  what  is  the  ratio  of  the 
cutting  to  the  return  speed  ? 

2.  On  a  36"  planer  the  ratio  of  the  cutting  speed  to  the  re- 
turn speed  of  the  table  is  1  to  2.94,  with  a  cutting  speed  of  68 
ft.  per  minute.    What  is  the  return  speed  ? 

3.  It  is  necessary  to  plane  a  bench  block  on  the  top  and 
bottom  surfaces.  An  equal  amount  of  stock  is  to  be  removed 
from  each  side,  and  the  thickness  of  the  casting  should  be  re- 
duced from  ly%  in.  to  If  in.  What  thickness  of  stock  should 
be  cut  from  the  top  surface  ? 

4.  A  piece  of  work  on  the  planer  is  10  in.  thick ;  it  is  re- 
duced to  a  certain  size  in  5  cuts ;  at  the  first  cut  the  tool  takes 
off  ^  in.  from  the  thickness ;  then  i  in.  ;  g\  in.,  ^i^-  in.,  and  the 
fifth  cut  is  ^ly  in.  What  is  the  thickness  of  the  finished 
piece  ? 

5.  On  a  cast  iron  block  6  in.  square^  a  groove  y\"  wide  and 
I"  deep  must  be  planed.  The  planer  makes  12  strokes  per 
minute  and  the  down  feed  is  set  at  -^"  for  every  stroke.  How 
long  will  it  take  to  cut  the  groove  ? 

Drilling  Machines.  —  Machines  for  drilling  holes  in  the  different  pieces 
made  in  a  machine  shop  are  divided  into  two  general  classes  —  vertical 
and  horizontal.  The  vertical  drilling  machines  include  those  with  a 
number  of  drill  spindles  called  multiple  spindles.  Besides,  there  are 
special  machines  of  both  classes,  as  portable  drills,  hand  drills,  etc. 

The  most  common  form  of  drill  is  the  vertical  drilling  machine  or 
drill-press.  The  machine  consists  of  a  frame  supporting  the  drill  spindle 
and  the  drilling  table,  and  an  arrangement  for  feeding  the  tool  into  the 
work  by  hand  or  power.     On  this  machine  the  work  to  be  drilled  is 

1  The  information  6"  square  is  not  necessary  for  the  solution  of  the 
problem,  but  is  given  to  add  interest  to  it.  The  same  thing  applies  to  some 
of  the  problems  in  drilling  on  the  following  pages,  where  the  size  of 
hole  is  given  but  not  required. 


PLANERS  AND  DRILLING   MACHINES  271 


Slwimq  Head  Drillino  Machine 


272  VOCATIONAL  MATHEMATICS 

placed  on  the  drilling  table,  and  is  held  stationary  by  means  of  a  clamp 
or  vise,  while  the  revolving  drill  is  fed  through  the  work  by  hand  or  by 
power.  The  feeding  mechanism  is  similar  to  that  of  the  lathe.  It  is 
more  convenient,  usually,  to  drill  small  work  in  a  speed  lathe  and  to  use 
the  drill-press  on  heavy  work. 

There  are  two  classes  of  drills  —  straight  and  twist ;  the  twist  drill  being 
the  latest  and  most  approved  in  the  leading  shops.  Whenever  a  manu- 
facturer is  making  standardized  pai-ts,  that  is,  uniform  parts  of  the  same 
machine,  day  after  day,  he  designs  and  builds  special  drilling  machines, 
also  special  milling  and  planing  machines,  gauges  called  jigs^  and  gauges 
to  produce  the  standard  or  duplicate  parts. 

Power  is  transmitted  from  the  pulley  on  the  shaft  to  the  countershaft 
at  the  bottom  of  the  machine.  Here  are  tight  and  loose  pulleys  so  that 
the  machine  may  be  stopped  by  shifting  the  belt  from  the  tight  to  the 
loose  pulley.  Different  materials  have  qualities  which  make  it  necessary 
to  use  different  cutting  speeds  in  order  to  remove  the  metal  quickly  and 
efficiently.  In  order  to  provide  for  this  a  cone  pulley  giving  a  number  of 
different  speeds  is  used.  The  cone  pulley  on  the  countershaft  is  the  same 
as  the  cone  pulley  on  the  headstock.  The  power  is  transmitted  from  the 
headstock  to  the  drill  spindle  by  bevel  gears.  The  power  is  transmitted 
from  the  headstock  shaft  to  the  feeding  spindle  —  which  in  turn  lets  the 
drill  spindle  down . 

Reamers.  —  Drills  cannot  be  relied  upon  to  make  holes  as  round, 
straight,  smooth,  or  uniform  in  diameter  as  are  required  in  the  construc- 
tion of  accurate  machinery.  To  make  these  holes  accurately  a  tool 
called  a  reamer  is  used.  It  has  two  or  more  teeth,  either  parallel  or  at 
an  angle  with  each  other.  The  periphery  of  a  drill  or  reamer  is  the  distance 
around  the  outside,  and  the  peripheral  speed  is  the  distance  a  point 
in  the  circumference  travels  in  a  minute.  It  is  equal  to  the  number  of 
turns  that  the  tool  makes  in  a  minute  multiplied  by  the  circumference. 
The  proper  speed  at  which  to  run  a  drill  depends  upon  the  kind  of  drill, 
its  size,  and  the  material  to  be  drilled.  A  large  drill  must  run  more 
slowly  than  a  smaller  one,  the  turns  per  minute  becoming  less  the  larger 
the  size  of  the  drill.  For  instance,  a  drill  \"  in  diameter  should  make 
twice  as  many  revolutions  per  minute  as  a  I"  drill  and  four  times  as 
many  as  one  2"  in  diameter,  used  on  the  same  material.  The  peripheral 
speeds  usually  recommended  for  carbon  steel  drills  are  as  follows : 

Wrought  iron  or  steel,  30ft.  per  min.  High  speed  drills,  60  to  70  ft.  per  min. 
Cast  iron,  35  ft.  per  min.  High  speed  drills,  60  to  80  ft.  per  min. 

Brass,  60  ft.  per  min.  High  speed  drills,  100  to  140 ft.  per  min. 


PLANERS  AND  DRILLING  MACHINES  273 

The  feed  of  a  drill  is  the  amount  the  drill  enters  the  hole  for  each  turn 
or  revolution,  and  in  good  practice  the  feed  may  be  set  to  give  1"  depth 
of  the  hole  for  every  96  to  125  revolutions,  according  to  the  size  of  the 
drill,  the  material  of  which  the  drill  is  made  and  the  kind  of  metal  drilled. 

Example.  —  How  long  will  it  take  a  one-inch  drill,  making 
134  R.  P.  M.,  with  a  feed  of  .012"  per  revolution,  to  drill  a 
hole  1^"  deep  in  cast  iron  ? 

— ^—  =  126  rev.  1  rev.  =  —  of  a  minute 

.012  134 

125  rev.  =  126  X  -^  =  —  =  .933  of  a  minute.    Ans. 
134      134 

Example.  —  A  piece  of  wrought  iron  2.69"  thick  is  to  have 
two  If"  holes  drilled  through  it.  If  the  drill  makes  112 
R.  P.  M.,  what  must   the  feed   be   to   drill  each  hole  in  two 

minutes  ? 

112  X  2  =  224  revs,  in  2  min. 
^i,  of  2.69"  =  .012"  feed.     Ans. 

Example.  —  A  2"  drill  is  used  in  drilling  an  electrical  gen- 
erator bed  plate.  If  the  drill  is  making  67  R.  P.  M.,  and  is 
being  fed  to  the  work  at  the  rate  of  .015"  per  revolution,  how 
deep  will  the  hole  be  when  the  drill  has  worked  4|  minutes  ? 

67  X  .016"  =  1.006"  depth  per  min. 
1.005x4.5=4.5225".     Ans. 

Example.  —  What  will  be  the  R.  P.  M.  of  a  drill  If"  in 
diameter  used  for  drilling  out  a  lathe  spindle  30.24"  long,  the 
feed  being  .015"  per  revolution,  and  the  time  given  the  mar 
chinist  to  do  the  job  being  42  minutes  ? 

80.24       OA1A  2016       Aan    r>   njr        a 

:5T5-  =  2016  .^  =  48R.P.M.     Ans. 

Example.  —  Conditions  being  the  same  as  in  the  above  ex- 
ample, what  length  of  a  spindle  could  be  drilled  in  50  minutes  ? 

48  X  .016  =  .720"  per  min. 
.720"  X  50  =  36"  in  60  min.     Ans. 


274  VOCATIONAL  MATHEMATICS 

EXAMPLES 

1.  How  long  will  it  take  a  carbon  steel  drill  to  drill  a  J" 
hole  through  a  piece  of  wrought  iron  l|f"  thick,  if  the  drill 
makes  105  E.  P.  M.,  with  the  feed  at  .011''  per  revolution  ? 

2.  With  the  feed  at  .015''  per  revolution  and  the  speed  at 
115  E.  P.  M.,  how  long  will  it  take  to  drill  a  hole  with  a  2" 
carbon  steel  drill  through  a  brass  bushing  4|"  long? 

3.  A  machinist  using  a  high  speed  drill  If"  in  diameter,  is 
to  drill  4  bolt  holes  in  the  base  of  an  electrical  generator,  the 
holes  to  be  2\"  long,  the  drill  to  make  164  R.  P.  M.,  with  the 
feed  at  .018"  per  revolution.  How  long  will  it  take  him  to  do 
the  job  if  3  minutes  are  allowed  for  the  setting  for  each  hole  ? 

4.  A  I"  drill  working  on  brass  is  running  at  306  E.  P.  M. 
and  advancing  at  the  rate  of  .015"  per  revolution.  How  long 
will  it  take  to  drill  through  a  piece  lyV'  thick  ? 

5.  The  E.  P.  M.  of  a  carbon  steel  drill  ly%"  in  diameter 
is  113,  and  the  feed  is  .013"  per  revolution.  How  much  time 
will  a  machinist  require  to  drill  18  holes  in  a  cylinder  head 
1^"  thick,  allowing  one  minute  for  the  setting  of  each  hole  ? 

6.  A  high  speed  drill  If"  in  diameter  is  advancing  at  the 
rate  of  .018"  per  revolution,  making  229  E.  P.  M.  while  drill- 
ing an  angle  iron  If"  thick.  How  long  will  it  take  the  drill  to 
go  through  it  ? 

7.  In  Example  6,  what  is  the  periphery  speed  in  feet  per 
minute  of  the  drill  ? 

8.  A  cast  iron  casing  for  a  steam  turbine  is  to  have  a  series 
of  holes  li"  in  diameter  drilled  in  it  2^"  deep  with  a  high 
speed  drill  making  250  E.  P.  M.  What  must  the  feed  be  per 
revolution  to  drill  each  hole  in  f  of  a  minute  ? 

9.  The  holes  in  a  face  plate  4"  thick  are  to  be  drilled 
with  a  1-in.  carbon  steel  drill  which  makes  134  E.  P.  M. 
(a)  What  feed  will  be  required  to  drill  through  the  plate  in 


PLANERS  AND  DRILLING  MACHINES  275 

2J  minutes?     (6)  What  will  be  the  periphery  speed  in  feet 
per  minute  of  the  drill  ? 

10.  What  must  be  the  R.  T.  M.  of  a  l.J"  drill,  feeding'  at 
the  rate  of  .013"  per  revolution,  to  drill  a  hole  3.'/'  deep  in  an 
engine  bed  in  3  minutes  ? 

11.  A  2"  drill  is  used  in  drilling  an  electrical  generator  bed 
platen.  If  the  drill  is  making  87  R.  P.  M.,  and  is  being  fed 
to  the  work  at  the  rate  of  .015"  per  revolution,  how  deep  will 
the  hole  be  when  the  drill  has  worked  4 J  minutes  ? 

12.  What  will  be  the  R.  P.  M.  of  a  1^"  carbon  drill  used  in 
drilling  out  an  engine  lathe  spindle  24.05"  in  24.GG  minutes  ? 
What  will  be  the  total  number  of  revolutions?  (Same  rate 
of  feed  as  in  Ex.  11.) 


PART   X  — TEXTILE   CALCULATIONS 

CHAPTER   XIX 

YARNS 

Worsted  Yarns.  —  All  kinds  of  yarns  used  in  the  manufacture 
of  cloth  are  divided  into  sizes  based  on  the  relation  between 
weight  and  length.  To  illustrate :  Worsted  yarns  are  made 
from  combed  wools,  and  the  size,  technically  called  the  counts,  is 


Roving  or  Yarn  Scales 
These  scales  will  weigh  one  pound  by  tenths  of  grains  or  one  seventy- 
thousandth  part  of  one  pound  avoirdupois,  rendering  them  well  adapted  for 
use  in  connection  with  yarn  reels,  for  the  numbering  of  yarn  from  the  weight 
of  hank,  giving  the  weight  in  tenths  of  grains  to  compare  with  tables. 

based  upon  the  number  of  lengths  (called  hanks)  of  560  yards 
required  to  weigh  one  pound.  Thus,  if  one  hank  weighs  one 
pound,  the  yarn  will  be  number  one  counts,  while  if  20  hanks 

276 


TEXTILE  CALCULATIONS 


277 


are  required  for  one  pound,  the  yarn  is  the  20's,  etc.  The 
greater  the  number  of  hanks  necessary  to  weigh  one  pound, 
the  higher  the  counts  and  the  finer  the  yarn.  The  hank,  or 
500  yai'ds,  is  the  unit  of  measurement  for  all  worsted  yarns. 

Lknotu  for  Worsted  Yarns 


No. 

Yards 
PRR  Lb. 

N.». 

Yabdh 
PKR  Lb. 

No. 

Yariw 
PER  Lb. 

No. 

Yardk 
PKR  Lh. 

1 
2 
3 
4 

560 
1120 
1680 
2240 

6 
6 

7 
8 

2800 
3360 
3920 
4480 

0 
10 
11 
12 

6040 
6600 
6160 
6720 

13 
14 
16 

16 

7280 
7840 
8400 
8960 

Woolen  Yams.  —  In  worsted  yams  the  fibers  lie  parallel  to 
each  other,  while  in  woolen  yarns  the  fibers  are  entangled. 
This  difference  is  due  entirely  to  the  different  methods  used 


Yarn  Reel 
For  reeling  and  measuring  lengths  of  cotton,  woolen,  and  worsted  yams. 

in  working  up  the  raw  stock.  In  woolen  yarns  there  is  a  great 
diversity  of  systems  of  grading,  varying  according  to  the  dis- 
tricts in  which  the  grading  is  done.    Among  the  many  systems 


278 


VOCATIONAL  MATHEMATICS 


are  the  English  skein,  which  differs  in  various  parts  of  Eng- 
land ;  the  Scotch  spyndle  ;  the  American  run  ;  the  Philadelphia 
cut;  and  others.  In  these  lessons  the  run  system  will  be  used 
unless  otherwise  stated.  This  is  the  system  used  in  New 
England.  The  run  is  based  upon  100  yards  per  ounce,  or 
1600  yards  to  the  pound.  Thus,  if  100  yards  of  woolen  yarn 
weigh  one  ounce,  or  if  1600  yards  weigh  one  pound,  it  is 
technically  termed  a  No.  1  run ;  and  if  300  yards  weigh  one 
ounce,  or  4800  yards  weigh  one  pound,  the  size  will  be  No.  3 
run.  The  finer  the  yarn,  or  the  greater  the  number  of  yards 
necessary  to  weigh  one  pound,  the  higher  the  run. 

Length  for  Woolen  Yarns  (Run  System) 


No. 

Taros 
PER  Lb. 

No. 

Yards 
PER  Lb. 

No. 

Yards 
PER  Lb. 

No. 

Yards 
PER  Lb. 

f 

200 
400 

800 
1200 

1 

If 

1600 
2000 
2400 
2800 

2 

^ 

2f 

3200 
3600 
4000 
4400 

3 

H 

4800 

5200 

.  5600 

Raw  Silk  Yarns.  —  For  raw  silk  yarns  the  table  of  weights 
is: 

16  drams  =  1  ounce 

16  ounces  =  1  pound 

256  drams  =  1  pound 

The  unit  of  measure  for  raw  silk  is  256,000  yards  per  pound. 
Thus,  if  1000  yards  —  one  skein  —  of  raw  silk  weigh  one 
dram,  or  if  256,000  yards  weigh  one  pound,  it  is  known  as 
1-dram  silk,  and  if  1000  yards  weigh  two  drams  the  yarn  is 
called  2-dram  silk,  hence  the  following  table  is  made : 

1-dram  silk  =  1000  yards  per  dram,  or  256,000  yards  per  lb. 
2-dram  silk  =  1000  yards  per  2  drams,  or  128,000  yards  per  lb. 
4-dram  silk  =  1000  yards  per  4  drams,  or  64,000  yards  per  lb. 


TEXTILE  CALCULATIONS 


279 


■ 

Dramr  i'br  1000  Yards 

Yards  prr  Pound 

Yardh  prr  Ounck 

1 

2r)<i,ooo 

in.ooo 

U 

204,800 

12,800 

n 

? 

? 

u 

I4rt,28r) 

9143 

2 

128,000 

8000 

H 

118,777 

7111 

^ 

102,400 

6400 

^ 

1)3,091 

6818 

8 

? 

? 

Si 

78,769 

4923 

^ 

73,143 

4571 

Linen  Yarns.  —  The  sizes  of  linen  yarns  are  based  on  the  lea 
or  cuts  per  pound  and  the  pounds  per  spindle.  A  cut  is  300 
yards  and  a  spindle  14,000  yards.  A  continuous  thread  of 
several  cuts  is  a  hank,  as  a  10-cut  hank,  which  is  10  x  300  = 
3000  yards  per  hank.  The  number  of  cuts  per  pound,  or  the 
leas,  is  the  number  of  the  yarn,  as  30's,  indicating  30  x  300  = 
9000  yards  per  pound.  Eight-pound  yarn  means  that  a  spindle 
weighs  8  pounds  or  that  the  yarn  is  6-lea  (14,400  -j-  8)  -r-  300  =  G. 

Cotton  Yams.  —  The  sizes  of  cotton  yarns  are  based  upon  the 
system  of  840  yards  to  1  hank.  That  is,  840  yards  of  cotton 
yarn  weighing  1  pound  is  called  No.  1  counts. 

Spun  Silk.  —  Spun  silk  yarns  are  graded  on  the  same  basis 
as  that  used  for  cotton  (840  yards  per  pound),  and  the  same 
rules  and  calculations  that  apply  to  cotton  apply  also  to  spun 
silk  yarns. 

Two  or  More  Ply  Yams.  —  Yarns  are  frequently  produced  in 
two  or  more  pi}- ;  that  is,  two  or  more  individual  threads  are 
twisted  together,  making  a  double  twist  yarn.  In  this  case 
the  size  is  given  as  follows : 

2/30's  means  2  threads  of  30's  counts  twisted  together,  and  3/30'8  would 
mean  3  threads,  each  a  30*8  counts,  twisted  together. 


280  VOCATIONAL  MATHEMATICS 

(The  figure  before  the  line  denotes  the  number  of  threads  twisted  to- 
gether, and  the  figure  following  the  line  the  size  of  each  thread.) 

Thus  when  two  threads  are  twisted  together,  the  resultant 
yarn  is  heavier,  and  a  smaller  number  of  yards  are  required  to 
weigh  one  pound. 

For  example  :  30's  worsted  yarn  ecjuals  16,800  yd.  per  lb.,  but  a  two- 
ply  thread  of  30's,  expressed  2/30's,  requires  only  8400  yards  to  the  pound, 
or  is  equal  to  a  15's;  and  a  three-ply  thread  of  oO's  would  be  equal  to  a 
lO's. 

When  a  yarn  is  a  two-ply,  or  more  than  a  two-ply,  and  made 
up  of  several  threads  of  equal  counts,  divide  the  number  of  the 
single  yarn  in  the  required  counts  by  the  number  of  the  ply, 
and  the  result  will  be  the  equivalent  in  a  single  thread. 

To  find  the  Weight  in  Grains  of  a  Given  Number  of  Yards  oj 
Worsted   Yarn  of  a  Known  Coiint 

Example.  —  Find  the  weight  in  grains  of  125  yards  of  20's 
worsted  yarns. 

No.  I's  worsted  yarn    =  560  yards 

No.  20 's  worsted  yarn  =  11,200  yards  to  1  lb. 

1  lb.  worsted  yarn        =  7000  grains 

125 


11,200 


If  11,200  yards  of  20's  worsted  yarn  weigh  7000'  grains,  then 
of  7000  =  -^  X  7000  =^  =  78.125  grains. 

1 1 ,  Z\)[j  o 

Note.  —  Another  method:  Multiply  the  given  number  of  yards  by 
7000,  and  divide  the  result  by  the  number  of  yards  per  pound  of  the 
given  count. 

125  X  7000  =  875,000 
1  pound  20's  =  11,200 
875,000  -^  11,200  =  78.125  grains.     Ans. 

To  find  the  Weight  in   Grains  of  a   Given  Number  of  Yards 
of  Cotton   Yaim  of  a  Known  Count 

Example.  — Find  the  weight  in  grains  of  80  yards  of  20's 
cotton  yarn. 


TEXTILE  CALCULATIONS  281 

No.  1  '8  cotton  =  840  yards  to  a  lb. 
No.  20'8  cotton  =  lfl,800  yards  to  a  lb. 
1  lb.  =  7000  grains 

7000 
1  yd.  20'8  cotton  =  ^^^^  grains 

7000  700 

80  yd.  20\s  cotton  =  , ,  ^^  x  80  =  --  =  33.33  grains.     Ans. 
lO,o0U  JI 

It  is  customary  to  solve  examples  that  occur  in  daily  practice 

by  rule. 

The  rule  for  the  preceding  example  is  as  follows  : 

Multiply  the  given  number  of  yards  by  7000  and  divide  the 

result  by  the  number  of  yards  per  pound  of  the  given  count. 

80  X  7000  =  500,000 
600,000  H-  (20  X  840)  =  33.33  grains.     Ana. 
Note.  — 7000  is  always  a  multiplier  and  840  a  divisor. 

To  find  the  weight  in  ounces  of  a  given  number  of  yards  of 
cotton  yarn  of  a  known  count,  multiply  the  given  number  of 
yards  by  16,  and  divide  the  result  by  the  yards  per  pound 
of  the  known  count. 

To  find  the  weight  in  pounds  of  a  given  number  of  yards 
of  cotton  yarn  of  a  known  count,  divide  the  given  number  of 
yards  by  the  yards  per  pound  of  the  known  count. 

To  find  the  weight  in  ounces  of  a  given  number  of  yards  of 
woolen  yarn  (run  system),  divide  the  given  niunber  of  yards 
by  the  number  of  runs,  and  multiply  the  quotient  by  100. 

Note.  —  Calculations  on  the  run  basis  are  much  simplified,  owing  to 
the  fact  that  the  standard  number  (1600)  is  exactly  100  times  the  number 
of  ounces  contained  in  1  pound. 

Example.  —  Find  the  weight  in  ounces  of  6400  yards  of 
5-run  woolen  yarn. 

6400  ^  (6  X  100)  =  12.8  oz.     Ana. 

To  find  the  weight  in  pounds  of  a  given  number  of  yards  of 
woolen  yam  (run  system)  the  above  calculation  may  be  used, 
and  the  result  divided  by  16  will  give  the  weight  in  pounds. 


282  VOCATIONAL  MATHEMATICS 

To  find  the  weight  in  grains  of  a  given  number  of  yards  of 
woolen  yarn  (run  system),  multiply  the  given  number  of  yards 
by  7000  (the  number  of  grains  in  a  pound)  and  divide  the  re- 
sult by  the  number  of  yards  per  pound  in  the  given  run,  and 
the  quotient  will  be  the  weight  in  grains. 

EXAMPLES 

1.  How  many  ounces  are  there  (a)  in  6324  grains  ?     (6)  in 
3^  pounds  ? 

2.  How  many  pounds  are  there  in  9332  grains  ? 

3.  How  many  grains  are  there  (a)  in  1681  pounds  ?    (b)  in 
2112  ounces  ? 

4.  Give  the  lengths  per  pound  of  the  following  worsted 
yarns:  (a)  41's;  (b)  55's ;  (c)  lOo's;  (d)  115's;  (e)  93's. 

5.  Give  the  lengths  per  pound  of  the  following  woolen  yarns 
(run  system)  :  (a)  Qi's ;  (b)  6's ;  (c)  19's ;  (d)  17's ;  (e)  li's. 

6.  Give  the  lengths  per  pound  of  the  following  raw  silk 
yarns:  (a)  li's;  (b)  3's;  (c)  3f 's ;  (d)  20's;  (e)  28's.      ' 

7.  Give  the  lengths  per  ounce  of  the  following  raw  silk 
yarns:   (a)  4i's;  (6)  6i's;  (c)  8's  ;  (d)  9's;  (e)  14's. 

8.  What  are  the  lengths  of  linen  yarns  per  pound :   (a)  8's ; 
(b)  25's;  (c)  32's;  (d)  28's;  (e)  45's  ? 

9.  What  are  the  lengths  per  pound  of  the  following  cotton 
yarns :   (a)  lO's ;  (6)  32's ;  (c)  54's ;  (d)  80's ;  (e)  160's  ? 

10.  What  are  the  lengths  per  pound  of  the  following  spun 
silk  yarns:  (a)  30's ;  (b)  45's;  (c)  38's;  (d)  29's;  (e)  42's  ? 

11.  Make  a  table  of  lengths  per  ounce  of  spun  silk  yarns 
from  I's  to  20's. 

12.  Find  the  weight  in  grains  of  144  inches  of  2/20's  worsted 
yarn. 

13.  Find  the  weight  in  grains  of  25  yards  of  3/30's  worsted 
yarn. 


TEXTILE  CALCULATIONS  283 

14.  Find  the  weight  in  ounces  of  24,000  yards  of  2/40's 
cotton  yarn. 

15.  Find  tlie  weight  in  pounds  of  2,840,000  yards  of  2/60's 
cotton  yarn. 

16.  Find  the  weight  in  ounces  of  650  yards  of  IJ-run  woolen 
yarn. 

17.  Find  the  weight  in  grains  of  80  yards  of  .]-run  woolen 
yarn. 

18.  Find  the  weight  in  pounds  of  04,000  yards  of  5-run 
woolen  yarn. 

Solve  the  following  examples,  first  by  analysis  and  then  by 
rule: 

19.  Find  the  weight  in  grains  of  165  yards  of  35*s  worsted. 

20.  Find  the  weight  in  grains  of  212  yards  of  40's  worsted. 

21.  Find  the  weight  in  grains  of  118  yards  of  25's  cotton. 

22.  Find  the  weight  in  grains  of  920  yards  of  18's  cotton. 

23.  Find  the  weight  in  pounds  of  616  yards  of  16J's  woolen. 

24.  Find  the  weight  in  grains  of  318  yards  of  184's  cotton. 

25.  Find  the  weight  in  grains  of  25  yards  of  30's  linen. 

26.  Find  the  weight  in  pounds  of  601  yards  of  60's  spun  silk. 

27.  Find  the  weight  in  grains  of  119  yards  of  118's  cotton. 

28.  Find  the  weight  in  grains  of  38  yards  of  64's  cotton. 

29.  Find  the  weight  in  grains  of  69  yards  of  39's  worsted. 

30.  Find  the  weight  in  grains  of  74  yards  of  40's  worsted. 

31.  Find  the  weight  in  grains  of  113  yards  of  IJ's  woolen. 

32.  Find  the  weight  in  grains  of  147  yards  of  IJ's  woolen. 

33.  Find  the  weight  in  grains  of  293  yards  of  S's  woolen. 

34.  Find  the  weight  in  grains  of  184  yards  of  16^'s  worsted. 

35.  Find  the  weight  in  grains  of  91  yards  of  44'8  worsted. 

36.  Find  the  weight  in  grains  of  194  yards  of  68's  cotton. 

37.  Find  the  weight  in  pounds  of  394  yards  of  180'8  cotton. 


284  VOCATIONAL  MATHEMATICS 

38.  Fiml  the  weight  in  pounds  of  (>12  yards  of  60's  cotton. 

39.  Find  the  weight  in  grains  of  118  yards  of  44's  linen. 

40.  Find  the  weight  in  pounds  of  315  yards  of  32's  linen. 

41.  Find  the  weight  in  grains  of  84  yards  of  25's  worsted. 

42.  Find  the  weight  in  grains  of  112  yards  of  20's  woolen. 

43.  Find  the  weight  in  grains  of  197  yards  of  16's  woolen. 

44.  Find  the  weight  in  grains  of  183  yards  of  18's  cotton. 

45.  Find  the  weight  in  grains  of  134  yards  of  28's  worsted 

46.  Find  the  weight  in  grains  of  225  yards  of  34's  linen. 

47.  Find  the  weight  in  pounds  of  369  yards  of  16's  spun  silk. 

48.  Find  the  weight  in  pounds  of  484  yards  of  18's  spun  silk. 

To  find  the  Size  or  the  Counts  of  Cotton  Yarn  of  Known 
Weight  and  Length 

Example.  —  Find  the  size  or  counts  of  84  yards  of  cotton 
yarn  weighing  40  grains. 

Since  the  counts  is  the  number  of  hanks  to  the  pound, 

^  X    84  =  14,700  yd.  in  1  lb. 

14,700  H-  840  =  17.5  counts.     Ans. 

Rule.  —  Divide  840  by  the  given  number  of  yards  ;  divide 
7000  by  the  quotient  obtained  ;  then  divide  this  result  by  the 
weight  in  grains  of  the  given  number  of  yards,  and  the  quotient 

will  be  the  counts. 

840  -  84  =  10 
7000  -f-  10  =  700 
700  -^  40  =  17.5  counts.     Ans. 

To  find  the  Ran  of  a    Woolen   TJiread  of  Knoivn  Length  and 

Weight 
Example.  — If  50  yards  of  woolen  yarn  weigh  77.77  grains, 
what  is  the  run  '/ 

1«00  ^  50  =  32 
7000  -f-  32  =  218.75 
218.75  -  77.77  =  2.812-run  yarn.     Ans. 


TEXTILE   CALCULATIONS  285 

Rule.  —  Divide  1600  (the  number  of  yards  per  pound  of  1- 
run  woolen  yarn)  by  the  given  number  of  yards ;  then  divide 
7000  (the  grains  per  pound)  by  the  quotient;  divide  this 
(juotient  by  the  given  weight  in  grains  and  the  result  will  be 
the  run. 

To  find  the  Weight  in  Ounces  for  a  Oiven  Number  of  Yards  of 
Worsted  Yarn  of  a  Known  Count 

Example.  —  What  is  the  weight  in  ounces  of  12,650  yards 
of  30's  worsted  yarn  ? 

12,650  X  16  =  202,400 
202,400  ^  16,800  =  12.047  oz.     Ans, 

Rule.  —  Multiply  the  given  number  of  yards  by  16,  and 
divide  the  result  by  the  yards  per  pound  of  the  given  count, 
and  the  quotient  will  be  the  weight  in  ounces. 

To  find  the  Weight  in  Pounds  for  a  Oiven  Number  of  Yards  of 
Worsted  Yai'n  of  a  Known  Count 

Example.  —  Find  the  weight  in  pounds  of  1 ,500,800  yards 
of  40's  worsted  yam. 

1,500,800  -4-  22,400  =  67  lb.     Ans. 

Rule. — Divide  the  given  number  of  yards  by  the  number 
of  yards  per  pound  of  the  known  count,  and  the  quotient  will 
be  the  desired  weight. 

EXAMPLES 

1.  If  108  inches  of  cotton  yarn  weigh  1.5  grains,  find  the 
counts. 

2.  Find  the  size  of  a  woolen  thread  72  inches  long  which 
weighs  2.5  grains. 

3.  Find  the  weight  in  ounces  of  12,650  yards  of  2/30's 
worsted  yarn. 

4.  Find  the  weight  in  ounces  of  12,650  yards  of  40's 
worste<l  yarn. 


286  VOCATIONAL  MATHEMATICS 

5.  Find  the  weight  in  pounds  of  1,500,800  yards  of  40's 
worsted  yarn. 

6.  Find  the  weight  in  pounds  of  789,600  yards  of  2/30's 
worsted  yarn. 

7.  What  is  the  weight  in  pounds  of  851,200  yards  of  3/60's 
worsted  yarn  ? 

8.  If  33,600  yards  of  cotton  yarn  weigh  5  pounds,  find  the 
counts  of  the  cotton. 

Buying  Yarn,  Cotton,  Wool,  and  Rags 

Ev^ery  fabric  is  made  of  yarn  of  definite  quality  and  quan- 
tity. Therefore,  it  is  necessary  for  every  mill  man  to  buy 
yarn  or  fiber  of  different  kinds  and  grades.  Many  small  mills 
buy  cotton,  wool,  yarn,  and  rags  from  brokers  who  deal  in 
these  commodities.  The  prices  rise  and  fall  from  day  to  day 
according  to  the  law  of  demand  and  supply.  Price  lists 
giving  the  quotations  are  sent  out  weekly  and  sometimes  daily 
by  agents  as  the  prices  of  materials  rise  or  fall.  The  follow- 
ing are  quotations  of  different  grades  of  cotton,  wool,  and 
shoddy,  quoted  from  a  market  list : 

QxTANTiTY  Price 

8103  lb.  white  yarn  shoddy  (best  all  wool) $0,485 

8164  lb.  white  knit  stock  (best  all  wool) 365 

2896  lb.  pure  indigo  blue 315 

1110  lb.  fine  dark  merino  wool  shoddy .225 

410  lb.  fine  light  merino  woolen  rags .115 

718  lb.  cloakings  (cotton  warp) 02 

872  lb.  wool  bat  rags ,     .         .035 

96  lb.  2/20's  worsted  (Bradford)  yarn 725 

408  lb.  2 /40's  Australian  yarn 1.35 

593  lb.  1/50's  delaine  yarn 1.20 

987  lb.  16  cut  merino  yarn  (50%  wool  and  50%  shoddy)   .     .     .          .285 
697  lb.  carpet  yam  60  yd.  double  reel  wool  filling 235 

Find  the  total  cost  of  the  above  quantities  and  grades  of 
textiles. 


TEXTILE  CALCULATIONS  287 


Alligation 

It  is  customary  in  mills  to  mix  different  fibers  at  different 
prices  in  order  to  make  a  product  of  some  intermediate  quality 
or  price.  The  process  of  finding  the  value  of  the  product  is 
called  alUgation. 

RuLK.  —  To  find  the  average  cost  per  pound  of  a  mixture, 
when  the  proportion  of  the  materials  mixed  and  their  prices 
are  given,  divide  the  total  value  of  the  materials  mixed  by 
the  sum  of  the  amounts  put  in,  and  the  quotient  will  be  the 
average  price  per  pound. 

A  mill  man  desires  to  find  the  average  value  per  pound  for 
the  following  lot  of  wool : 

218  lb.  at    81  cts.  per  lb.  218  x    .81  =  $   176.68 

413  lb.  at    79  cts.  per  lb.  413  x    .79  =      362.27 

284  lb.  at   82  cts.  per  lb.  284  x    .82  =       232.88 

264  lb.  at    83  cts.  per  lb.  264  x    .83  =       219.12 

18  lb.  at  103  cts.  per  lb.  _J8  x  1.03  =         19.44 

1197  1b.         :=.«;  1074.29 

1074.29  ^  1197  lb.  =  .$0,897,  price  of  mixture  per  pound. 

EXAMPLES 

Find  the  average  value  per  pound  of  the  following  lots  of 
wool : 

1.  217  1b.  at  $0.82  4.  164  lb.  at  $0.94 
384  lb.  at  .86  218  lb.  at  .81 
295  lb.  at       .89  98  lb.  at     1.08 

2.  198  lb.  at  S0.78  5.  208  lb.  at  $0.81 
164  lb.  at  .81  161  lb.  at  .78 
312  lb.  at       .74  94  lb.  at     1.12 

a   217  lb.  at  $0.94  6.   174  lb.  at  $0.79 

108  lb.  at       .78  101  lb.  at        .91 

79  lb.  at     1.12  68  lb.  at      1.08 


112  lb. 

at 

$0.74 

184  lb. 

at 

.88 

101  lb. 

at 

1.02 

211  lb. 

ai3 

$0.78 

191  lb. 

at 

.91 

294  lb. 

at 

1.11 

198  lb. 

at 

$0.75 

208  lb. 

at 

.88 

69  1b. 

at 

1.03 

69  1b. 

at 

$0.91 

511b. 

at 

.98 

288  VOCATIONAL  MATHEMATICS 

7. 


10. 


Alligation  Alternate 

The  process  of  finding  the  quantities  of  different  values  re- 
quired to  produce  a  mixture  of  a  given  value  is  called  alligatioyi 
alternate. 

A  mill  man  often  desires  to  find  the  amount  of  each  kind  of 
wool  that  must  be  mixed  to  produce  a  mixture  of  a  definite 
amount. 

Example.  —  How  much  wool  of  each  kind  at  respective 
values  of  81,  85,  and  96  cts.  must  be  mixed  to  produce  a 
mixture  of  89  cts.  per  pound  ? 


%Q{ 


81 


96 


+8x1=8 
+  4  X  1=^ 

12  gain 
-7  X  If  =12  loss 
Place  the  price  of  the  mixture  on  the  left  and  the  prices  of  the  ingre- 
dients on  the  right  of  the  brace.  The  differences  are  placed  on  the  right 
of  another  brace  as  plus  or  minus  as  shown  above.  One  pound  of  the  first 
and  1  of  the  second  will  give  a  gain  of  12  over  the  desired  mixture  ;  If  lb. 
of  the  third  will  make  up  the  difference  of  12. 

One  pound  of  the  first,  one  of  the  second,  and  If  pounds  of  the  third 
will  give  the  desired  mixture. 

EXAMPLES 

1.  How  much  wool  of  each  kind  is  required  at  the  respective 
values  of :  83,  87,  and  94  to  produce  a  mixture  of  88  cts.  ? 

2.  How  much  wool  of  each  kind  is  required  at  the  respective 
values  of :  78,  83,  and  88  to  produce  a  mixture  of  86  cts.  ? 

3.  How  much  wool  of  each  kind  is  required  at  the  respective 
values  of :  74,  84,  and  91  to  produce  a  mixture  of  87  cts.  ? 


TEXTILE  CALCULATIONS  289 

4.  How  much  wool  of  each  kind  is  required  at  the  respective 
values  of :  79,  86,  and  94  to  produce  a  mixture  of  88  cts.  ? 

5.  How  mucli  wool  of  each  kind  is  required  at  the  respective 
values  of :  78,  88,  and  9'^  to  produce  a  mixture  of  91  cts.  ? 

Example.  —  A  manufacturer  has  432  lb.  of  wool  of  a  value 
94  cts.  on  hand  which  he  desires  to  use  in  producing  a  lot 
worth  82  cts.  per  lb.  He  has  a  large  lot  of  a  cheaper  grade  of 
wool  (2587  lb.  worth  75  cts.).  How  much  of  this  cheaper 
grade  must  be  used  to  produce  a  mixture  worth  82  cts.  per 
pound  ? 


82 


04 


[76 


-  12  X  432  =  6184  loss 


+  7  X  740|  =  5184  gain 

He  should  mix  432  lb.  of  94  cts.  and  740f  lb.  of  76  cts.  wool  in 
order  to  produce  a  mixture  of  82  cts. 

Proof.— \U  X  4.32    lb.  =  §406.08 
75  X  740S  lb.  =     555.42^ 
11721  lb.  =§961.50^ 
1  lb.  =  82  cts. 

EXAMPLES 

1.  With  merino  wool  at  $  1.38  per  lb.  and  Australian  wool 
at  $  1.25  and  S  1.31  per  lb.,  what  proportion  of  each  should  be 
used  to  produce  a  mixture  costing  $  1.32  per  lb.  ? 

2.  How  will  four  kinds  of  cotton  fiber  worth  37  cents,  42 
cents,  43J  cents,  and  82  cents  per  pound  be  mixed  to  give  off 
an  intermediate  grade  of  cotton  to  be  valued  at  50  cents  per 
pound  ? 

3.  When  wool  is  valued  at  $1.33^  per  lb.  and  cotton  at 
87  cents,  how  much  of  each  must  be  used  to  produce  a  grade 
of  mixture  at  $  1.02  per  lb.  ? 

4.  Find  the  average  cost  per  lb.  of  a  mixture  made  up  as 
follows:  J  at  S  1.23  per  lb.,  J  at  87  cents  per  lb.,  and  the 
remainder  at  97  cents  per  lb. 


290  VOCATIONAL  MATHEMATICS 

5.  What  is  the  cost  per  lb.  of  the  following  mixture: 
I  Egyptian  at  $  1.02  per  lb.,  ^  upland  at  39  cents  per  lb.,  and 
J  Indian  cotton  at  28  cents  per  lb.  ? 

6.  In  a  certain  piece  of  cloth  two  woolen  yarns  of  different 
grades  and  a  silk  thread  are  used ;  j\  of  the  yarn  is  silk  and 
equal  amounts  of  the  two  grades  of  wool  are  used.  The  silk 
cost  $  2.25  per  lb.  and  the  wool  used  weighed  750  lb.  and  was 
valued  at  $  1.23  and  $  1.49  respectively.  How  much  silk  and 
wool  was  used  ?     What  was  the  cost  of  the  yarn  ? 

7.  How  many  bales  of  cotton,  the  average  weight  being 
490  lb.  to  a  bale,  would  be  required  to  supply  a  picker  room 
having  12  finisher  pickers,  each  machine  producing  11,250 
pounds  of  finished  laps  per  week,  the  loss  during  the  process 
being  3i  %  ? 

8.  How  many  bales  of  cotton,  the  average  weight  being 
495  lb.  to  a  bale,  would  be  required  to  supply  a  picker  room 
having  9  finisher  pickers,  each  machine  producing  10,895 
pounds  of  finished  laps  per  week,  the  loss  during  the  process 
being  3}  %  ? 

9.  How  many  bales  of  cotton,  the  average  weight  being 
500  lb.  to  a  bale,  would  be  required  to  supply  a  picker  room 
having  20  finisher  pickers,  each  machine  producing  11,250 
pounds  of  finished  laps  per  week,  the  loss  during  the  process 
being  4  %  ? 

10.  How  many  bales  of  cotton  would  be  required  for  an 
order  of  85,000  pounds  of  yarn,  the  average  weight  per  bale 
being  485  lb.  ?  The  loss  in  the  picker  room  is  2J%,  the  loss 
at  the  cards  is  3|%,  and  the  loss  is  3  %  (including  invisible 
loss)  during  the  remaining  processes. 


APPENDIX 

METRIC  SYSTEM 

The  metric  system  is  used  in  nearly  all  the  countries  of 
Continental  Europe  and  among  scientific  men  as  the  standard 
system  of  weights  and  measures.  It  is  based  on  the  meter  as 
the  unit  of  length.  The  meter  is  supposed  to  be  one  ten- 
millionth  part  of  the  length  of  the  meridian  j)assing  from  the 
equator  to  the  poles.  It  is  equal  to  about  39.37  inches.  The 
unit  of  weight  is  the  gram^  which  is  equal  to  about  one 
thirtieth  of  an  ounce.  The  unit  of  volume  is  the  liter',  which 
is  a  little  larger  than  a  quart. 

Measures  of  Length 

10  millimeters  (mm.)     .     .     .     .  =  1  centimeter cm. 

10  ceiitimetei-s =1  decimeter dm. 

10  decimeters =1  meter m. 

10  meters =1  dekameter Dm. 

10  dekameters =1  hektometer Urn. 

10  hektometers =1  kilometer Km. 

Measures  of  Surface  (not  Land) 

100  square  millimeters  (mm.)  .  .  =  1  squai*e  centimeter  .  .  sq.  cm. 
100  square  centimeters  .  .  •  .  =  1  square  decimeter  .  .  sq.  dm. 
10<3  wiuare  decimeters     .     .     .     .     =  1  square  meter sq.  m. 

Measures  of  Volume 

1000  cubic  millimeters  (mm.)  .  .  =  1  cu.  centimeter  .  .  .  cu.  cm. 
1000  cubic  centimeters  .  .  .  .  =  1  cubic  decimeter  .  .  .  cu.  dm. 
1000  cubic  decimeters     .     .     .     .     =  1  cubic  meter cu.  m. 

I  Hie  jrram  is  the  weight  (►f  one  cubir  centimeter  of  pure  di.stille<i  water  at 
a  temperature  of  ;i9.2°F. ;  the  kilogram  is  the  weight  of  1  liter  of  water;  the 
metric  ton  is  the  weight  of  1  cubic  meter  of  water. 

291 


292 


VOCATIONAL   MATHEMATICS 


Measures  of  Capacity 

10  milliliters  (ml.)     .....=  1  centiliter 

10  centiliters     .«....,=  1  deciliter 

10  deciliters =1  liter  i 

10  liters =1  dekaliter 

10  dekaliters =1  hektoliter 

10  hektoliters =  1  kiloliter 


Measures  of  Weight 


10  milligrams  (mg.) 
10  centigrams    .     . 


cl 
dl. 
1. 
Dl. 
HI. 
Kl. 


=  1  centigram eg 

=  1  decigram dg 


10  decigrams =1  gram 


10  grams 
10  dekagrams 
10  hektograms 
1000  kilograms 


=  1  dekagram   ......  Dg 

=  1  hektogram Hg 

=  1  kilogram Kg 

=  1  ton T 


Metric  Equivalent  Measures 

Measures  of  Length 

1  meter  =  39.37  in.  =  3.28083  ft.  =  1.0936  yd. 

1  centimeter  =      .3937  inch 

1  millimeter  =      .03937  inch,  or  ^^  inch  nearly 

1  kilometer  =      .62137  mile 

1  foot  =      .3048  meter 

1  inch  =    2.54  centimeters  =  25.4  millimeters 

Measures  of  Surface 

1  square  meter  =  10.764  sq.  ft.  =  1.196  sq.  yd. 

1  square  centimeter  =      .155  sq.  in. 

00155  sq.  in. 

836  square  meter 

,0929  square  meter 

.452  square  centimeters =645. 2  square  millimeters 


1  square  millimeter  = 
1  square  yard 
1  square  foot 
1  square  inch 


Measures  of  Volume  and  Capacity 
=  35.314  cu.  ft.  =  1.308  cu.  yd.  =  264.2  gal. 


1  cubic  meter 

1  cubic  decimeter     =  61.023  cu.  in.  =  .0353  cu 

1  cubic  centimeter    =      .061  cu.  in. 


ft. 


1  The  liter  is  equal  to  the  volume  occupied  by  1  cubic  decimeter. 


METRIC   SYSTEM  293 

1  liter  =  1  cubic  deoiineter  =  01.023  cu.  in.  =  .0363  cu.  ft.  = 

1.0607  <iimrt«   (U.  S.)=.2042  gallon  (U.  S.)  = 
2.202  lb.  of  water  at  02°  F. 

1  cubic  yaixl  =      .7(W6  cubic  meter 

1  cubic  foot  =      .02832  cubic  meter  =  28.317  cubic  decimeters  = 

28.317  liters 
1  cubic  inch  =  1(5.387  cubic  centimeters 

1  gallon  (British)      =    4.643  liters 
1  gallon  (U.  S  )        =    3.786  liters 

Measures  of  Weight 

1  i,Tam  =  15.432  grains 

1  kilogram  =    2.2045  pounds 

1  metric  ton  =      .5)842  ton  of  2240  lb,  r=  1U.08  cwt.  =  2204.0  lb. 

1  grain  =      .0048  gram 

1  ounce  avoirdupois  =  28.36  grams 

1  pound  =      .453(J  kilogram 

1  ton  of  2240  lb.       =    1.010  metric  tons  =  1010  kilograms 

Miscellaneous 

1  kilogram  per  meter  =  .0720  pound  i)er  foot 

1  gram  per  stjuare  millimeter  =  1.422  pounds  per  8<iuare  inch 

1  kilogram  per  scjuare  meter  =  .2084  pound  per  square  foot 

1  kilogram  per  cubic  meter  =  .0024  pound  per  cubic  foot 

1  degree  centigrade  —  1.8  degrees  Fahrenheit 

1  pound  per  foot  =  1.488  kilograms  per  meter 

1  pound  per  square  foot  =  4.882  kilograms  per  square  meter 

1  pound  per  cubic  foot  =  10.02  kilograms  per  cubic  meter 

1  degree  Fahrenheit  =  .6560  degree  centigrade 

1  Calorie  (French  Thermal  Unit)  =  3.9(58  B.  T.  U.  (British  Thermal  Unit) 

1  horse  power  =  33,0(K)  foot  pounds  per  minute  =  740  watts 

1  watt  (Unit  of  Electrical  Power)  =  .00134  horse  power  =  44.24  foot 

pounds  per  minute 
1  kilowatt  =  1000  watts  =  1.34  horse  power =44,240  foot  pounds  per  minute 

Table  of  Metric  Conversion 

To  change  meters  to  feet multiply  by  3,28088 

feet  to  meters multiply  l)y         .3048 

square  feet  U)  square  meters       .     .     .  multiply  by         .0029 

square  meters  to  square  feet       .     .     .  multiply  by  10.764 


294  VOCATIONAL  MATHEMATICS 

To  change  square  centimeters  to  square  inches    .  multiply  by        .155 

square  inches  to  square  centimeters    .  multiply  by  6.452 

inches  to  centimeters multiply  by  2.54 

centimeters  to  inches multiply  by        .3937 

grams  to  grains multiply  by  15.43 

grains  to  grams multiply  by         .0648 

grams  to  ounces multiply  by        .0358 

ounces  to  grams multiply  by  28.35 

pounds  to  kilograms multiply  by        .4536 

kilograms  to  pounds multiply  by  2.2045 

liters  to  quarts multiply  by  1.0567 

liters  to  gallons multiply  by        .2642 

gallons  to  liters multiply  by  3.78543 

liters  to  cubic  inches multiply  by  61.023 

cubic  inches  to  cubic  centimeters    .     .  multiply  by  16.387 

cubic  centimeters  to  cubic  inches    .     .  multiply  by        .061 

cubic  feet  to  cubic  decimeters  or  liters  multiply  by  28.317 

kilowatts  to  horse  power multiply  by  1.34 

calories  to  British  Thermal  Units   .     .  multiply  by  3.968 


EXAMPLES 

1.  Change  8  m.  to  centimeters  ;  to  kilometers. 

2.  Reduce  4  Km.,  6  m.,  and  2  m.  to  centimeters. 

3.  How  many  square  meters  of  carpet  will  cover  a  floor 
which  is  25.5  feet  long  and  24  feet  wide  ? 

4.  (a)  Change  6.5  centimeters  into  inches. 

(b)  Change  48.3  square  centimeters  into  square  inches. 

5.  A  cellar  18  m.  x  37  m.  x  2  m.  is  to  be  excavated ;  what 
will  it  cost  at  13  cents  per  cubic  meter  to  do  the  work  ? 

6.  How   many  liters  of   capacity  has    a   tank    containing 
5.2  cu.  m.  ? 

7.  What  is  the  weight  in  grams  of  31  cc.  of  water  ? 

8.  Give  the  approximate  value  of  36  millimeters  in  inches. 

9.  Change  84.9  square  meters  into  square  feet. 
10.    Change  23.6  liters  to  cubic  inches. 


METRIC  SYSTEM  295 

11.  If  a  tank  is  7.6  m.  by  1.4  m.  by  2.2  m.,  how  many  kilo- 
grams of  water  will  it  hold  ? 

12.  A  blue  print  of  a  t^ble  top  measures  .3.8  m.  by  78.6  em., 
what  are  its  length  and  width  in  inehes  ? 

J.3.    A  package  weighs  28  kg.    What  is  its  weight  in  pounds? 

14.  A  box  weighs  82.5  g.    How  many  ounces  does  it  weigh  ? 

15.  (a)  Change  14.(5  kilograms  to  pounds. 

(b)  Change  6.8  kilowatts  to  horse  power. 

(c)  Change  20.4  liters  to  quarts. 

(d)  Change  83.2  liters  to  gallons. 

(e)  Change  10.25  inches  to  centimeters. 

16.  A  cubic  foot  bf  water  weighs  about  62.5  lb.  Find  the 
weight  in  kilograms. 

17.  (a)  Change  30.5  sq.  in.  to  sq.  cm. 
(6)  Change  84.9  sq.  ft.  to  sq.  m. 

(c)  Change  8.5  gal.  to  liters. 

(d)  Change  164.3  grains  to  grams. 

(e)  Change  212.8  oz.  to  grams. 

18.  The  dimensions  of  a  meal  chest  on  a  foreign  blue  print 
are  2.6  m.,  .48  m.,  and  59.3  cm.  How  many  liters  of  meal  will 
it  hold? 

19.  Change  6.1  pounds  into  kilogi-ams. 

20.  A  rectangular  watering  trough  is  166  cm.  long,  58  cm. 
wide,  and  422  mm.  deep.  If  it  is  full  of  water,  what  are  its 
contents  in  liters  ? 


GRAPHS 

A  SHEET  of  paper,  ruled  with  horizontal  and  vertical  lin-es 
that  are  equally  distant  from  each  other,  is  called  a  sheet  of 
cross-section,  or  coordinate,  paper.  Every  tenth  line  is  very 
distinct  so  that  it  is  easy  for  one  to  measure  off  the  horizontal 
and  vertical  distances  without  the  aid  of  a  ruler.     Ruled  or 


Graph  showing  the  Variation  in  Price  of  Cotton  Yarn  for  a 
Series  of  Years 


coordinate  paper  is  used  to  record  the  rise  and  fall  of  the 
price  of  any  commodity,  or  the  rise  and  fall  of  the  barometer 
or  thermometer. 

Trade  papers  and  reports  frequently  make  use  of  coordinate 
paper  to  show  the  results  of  the  changes  in  the  price  of  com- 
modities.    In  this  way  one  can  see  at  a  glance  the  changes 

296 


GRAPHS  297 

and  condition  of  a  certain  commodity,  and  can  compare  these 
with  the  results  of  years  or  months  ago.  He  also  can  see 
from  the  slope  of  the  curve  the  rate  of  rise  or  fall  in  price. 

If  similar  comnuxlities  are  plotted  on  the  same  sheet,  the 
effect  of  one  on  the  other  can  be  noted.  Often  experts  are 
able  to  prophesy  with  some  certainty  the  price  of  a  commodity 
for  a  month  in  advance.  The  two  quantities  which  must  be 
employed  in  this  comparison  are  time  and  value,  or  terms 
corresponding  to  them. 

The  lowest  left-hand  corner  of  the  squared  paper  is  generally 
used  as  an  initial  point,  or  origin,  and  is  marked  0,  although 
any  other  corner  may  be  used.  The  horizontal  line  from  this 
corner,  taken  as  a  line  of  reference  or  axis,  is  called  the  ab- 
scissa. The  vertical  line  from  this  corner  is  the  other  axis, 
and  is  called  the  ordinate. 

Equal  distances  on  the  abscissa  (horizontal  line)  represent 
definite  units  of  time  (hours,  days,  months,  years,  etc.),  while 
equal  distances  along  the  ordinate  (vertical  line)  represent 
certain  units  of  value  (cost,  degrees  of  heat,  etc.). 

By  plotting,  or  placing  points  which  correspond  to  a  certain 
value  on  each  axis  and  connecting  these  points,  a  line  is  ob- 
tained that  shows  at  every  point  the  relationship  of  the  line 
to  the  axis. 

EXAMPLES 

1.  Show  the  rise  and  fall  of  temperature  in  a  day  from 
8  A.M.  to  8  P.M.,  taking  readings  every  hour. 

2.  Show  the  rise  and  fall  of  temperature  at  noon  every  day 
for  a  week. 

3.  Obtain  stock  quotation  sheets  and  plot  the  rise  and  fall 
of  cotton  for  a  week. 

4.  Show  the  rise  and  fall  of  the  price  of  potatoes  for  two 
months. 

5.  Show  a  curve  giving  the  amount  of  coal  used  each  day 
for  a  week. 


FORMULAS 

Most  technical  books  and  magazines  contain  many  formulas. 
The  reason  for  this  is  evident  when  we  remember  that  rules 
are  often  long  and  their  true  meaning  not  comprehended  until 
they  have  been  reread  several  times.  The  attempt  to  abbre- 
viate the  length  and  emphasize  the  meaning  results  in  the 
formula,  in  which  whole  clauses  of  the  written  rule  are  ex- 
pressed by  one  letter,  that  letter  being  understood  to  have 
throughout  the  discussion  the  same  meaning  with  which  it 
started. 

To  illustrate  :  One  of  the  fundamental  laws  of  electricity  is  that  the 
quantity  of  electricity  flowing  through  a  circuit  (flow  of  electricity)  is 
equal  to  the  quotient  (expressed  in  amperes)  obtained  by  dividing  the 
electric  motive  force  (pressure,  or  expressed  in  volts,  voltage)  of  the 
current  by  the  resistance  (expressed  in  ohms). 

One  unfamiliar  with  electricity  is  obliged  to  read  this  rule  over  several 
times  before  the  relations  between  the  different  parts  are  clear.  To  show 
how  the  rule  may  be  abbreviated, 

Let  I  =  quantity  of  electricity  through  a  wire  (amperes) 
E  =  pressure  of  the  current  (volts) 
B  =  resistance  of  the  current  (ohms) 

Then  7=  E ^  R  =- 
B 

It  is  customary  to  allow  the  first  letter  of  the  quantity  to  represent  it  in 
the  formula,  but  in  this  case  I  is  used  because  the  letter  G  is  used  in  an- 
other formula  with  which  this  might  be  confused. 

Translating  Rules  into  Formulas 

The  area  of  a  trapezoid  is  equal  to  the  sum  of  the  two  parallel 
sides  multiplied  by  one  half  the  perpendicular  distance  between 
them. 

298 


FORMULAS  299 

We  may  abbreviate  this  rule  by  letting 

A  =  area  of  trapezoid 
L  =  length  of  longest  parallel  side 
M  =  length  of  shortest  parallel  side 
J\r=  length  of  perpendicular  distance  between  them 

Then  A  =  {L  -\-  M)  x^,  or 

The  area  of  a  circle  is  equal  to  the  square  of  the  radius 
multiplied  by  3.1416.  When  a  number  is  used  in  the  formula 
it  is  called  a  constant,  and  is  sometimes  represented  by  a  ieiMf. 
In  this  case  3.1416  is  represented  by  the  Greek  letter  n  (pi). 

Let  A  =  area  of  circle 
R  =  radius  of  circle 
Then  A  =  ir  x  Ji^,  or  (the  multiplication  sign  is  usually  left  out  between 
letters) 

A  =  TRi 

Thus  we  see  that  a  formula  is  a  short  and  simple  way  of 
stating  a  rule.  Any  formula  may  be  written  or  expressed  in 
words  and  is  then  called  a  rule.  The  knowledge  of  formulas 
and  of  their  use  is  necessary  for  nearly  every  one  engaged  in 
the  higher  forms  of  mechanical  or  technical  work. 

*  When  two  or  more  quantities  are  to  be  multiplied  or  divided  or  other- 
wise operated  upon  by  the  same  quantity,  they  are  often  grouped  together 
by  means  of  parentheses  (  )  or  braces  {  },  or  brackets  [  ].  Any  number 
or  letter  placed  before  or  after  one  of  these  parentheses,  with  no  other 
sign  between,  is  to  multiply  all  that  is  grouped  within  the  parentheses. 

V 
In  the  trapezoid  case  above,  —  is  to  multiply  the  sum  of  L  and  Jf,  hence 

the  parentheses.     To  prevent  confusion,  different  signs  of  aggi-egation 
may  be  used  for  different  combinations  in  the  same  problem. 
For  instance, 

V=  lirll\f(r^  +  r'-^)  +  —  1  which  equals 
3       L  2  2  J 

V=iirH(r^+  r'2)  +  iir//» 


300  VOCATIONAL  MATHEMATICS 

EXAMPLES 
Abbreviate  the  following  rules  into  formulas : 

1.  One  electrical  horse  power  is  equal  to  746  watts. 

2.  One  kilowatt  is  equal  to  1000  watts. 

3.  The  number  of  watts  consumed  in  a  given  electrical  cir- 
cuit, such  as  a  lamp,  is  obtained  by  multiplying  the  volts  by 
the  amperes. 

4.  The  volts  equal  the  watts  divided  by  the  amperes. 

5.  Amperes  equal  the  watts  divided  by  the  volts. 

6.  To  find  the  pressure  in  pounds  per  square  inch  of  a 
column  of  water,  multiply  the  height  of  the  column  in  feet  by 
.434. 

7.  To  find  the  diameter  of  a  pump  cylinder  required  to 
move  a  given  quantity  of  water  per  minute  (100  ft.  of  piston 
being  the  standard  of  speed),  divide  the  number  of  gallons  by 
4,  then  extract  the  square  root,  and  the  quotient  will  be  the 
diameter  in  inches  of  the  pump  cylinder. 

8.  To  find  quantity  of  water  elevated  in  one  minute  (run- 
ning at  100  feet  of  piston  per  minute),  square  the  diameter  of 
the  water  cylinder  in  inches  and  multiply  by  4. 

9.  To  find  the  horse  power  necessary  to  elevate  water  to  a 
given  height,  multiply  the  weight  of  the  water  elevated  per 
minute  in  pounds  by  the  height  in  feet,  and  divide  the  prod- 
uct by  33,000.  (An  allowance  should  be  added  for  water  fric- 
tion, and  a  further  allowance  for  loss  in  steam  cylinder,  say 
from  20  to  30  per  cent.) 

10.  In  a  steam  pump  the  effective  pressure  equals  the  area  of 
the  steam  piston,  multiplied  by  the  steam  pressure,  minus  the 
area  of  the  water  piston,  multiplied  by  the  pressure  of  water 
per  square  inch,  which  gives  the  resistance.  (A  margin  must 
be  made  between  the  power  and  the  resistance  to  move  the 
pistons  at  the  required  speed  —  say  from  20  to  40  per  cent,  ac- 
cording to  speed  and  other  conditions.) 


FORMULAS  301 

11.  To  find  the  capacity  of  a  cylinder  in  gallons.  Multi- 
plying the  area  in  inches  by  the  length  of  stroke  in  inches 
will  give  the  total  number  of  cubic  inches;  divide  this  amount 
by  231  (which  is  the  cubical  contents  of  a  United  States  gal- 
lon in  inches)  and  the  product  is  the  capacity  in  gallons. 

12.  To  find  the  length  of  an  arc  of  a  circle :  Multiply  the 
diameter  of  the  circle  by  the  number  of  degrees  in  the  arc  and 
this  product  by  .0087266. 

13.  To  find  the  area  of  a  sector  of  a  circle :  Multiply  the 
number  of  degrees  in  the  arc  of  the  sector  by  the  square  of  the 
radius  and  by  .008727 ;  or,  multiply  the  arc  of  the  sector  by 
half  its  radius. 

Translating  Formulas  into  Rules 

In  order  to  understand  a  formula,  it  is  necessary  to  be  able 
to  express  it  in  simple  language. 

1.  One  of  the  simplest  formulas  is  that  for  finding  the  area 
of  a  circle,  A  =  7r  K^ 

Here  A  stands  for  the  area  of  a  circle, 

B  for  the  radius  of  the  circle. 

T  is  a  constant  quantity  and  is  the  ratio  of  the  circumference  of  a 
circle  to  its  diameter.  The  exact  value  cannot  be  expressed  in  figures, 
but  for  ordinary  purposes  is  called  3.1416  or  3^. 

Therefore,  the  formula  reads,  the  area  of  a  circle  is  equal  to 
the  square  of  the  radius  multiplied  by  3.1416. 

2.  The  formula  for  finding  the  area  of  a  rectangle  is 

A  =  Lx  W 

Here  A  =  area  of  a  rectangle 
L  =  length  of  rectan^e 
ir  =  width  of  rectangle 

The  area  of  a  rectangle,  therefore,  is  found  by  multiplying 
the  length  by  the  width. 


302  VOCATIONAL   MATHEMATICS 

EXAMPLES 

Express  the  facts  of  the  following  formulas  as  rules : 

1.  Electromotive  force  or  voltage  of  electricity  delivered  by 
a  current,  when  current  and  resistance  are  given : 

E  =  RI 

2.  For  the  circumference  of  a  circle,  when  the  length  of  the 
radius  is  given: 

C^2nR0Y  ttD 

3.  For  the  area  of  an  equilateral  triangle,  when  the  length 
of  one  side  is  given:  a2^'3 

4.  For  the  volume  of  a  circular  pillar,  when  the  radius  and 
height  are  given : 

V=  TrR'h 

5.  For  the  volume  of  a  square  pyramid,  when  the  height 
and  one  side  of  the  base  are  given : 

~    3 

6.  For  the  volume  of  a  sphere,  when  the  diameter  is  given : 

6 

7.  For  the  diagonal  of  a  rectangle,  when  the  length  and 
breadth  are  given : 

8.  For  the  average  diameter  of  a  tree,  when  the  average 
girth  is  known  '•  j^_(^ 

TT 

9.  For  the   diameter  of   a  ball,  when  the  volume  of  it  is 
known. 

ir 


FORMULAS  303 

10.  The  diameter  of  a  circle  may  be  obtained  from  the  area 
by  the  following  formula : 

/>  =  1.1283  xV]4 

11.  The  number  of  miles  in  a  given  length,  expressed,  in 
feet,  may  be  obtained  from  the  formula 

3f=.00019xF 

12.  The  number  of  cubic  feet  in  a  given  volume  expressed 
in  gallons  may  be  obtained  from  the  formula 

0=. 13367  X  Q 

13.  Contractors  express  excavations  in  cubic  yards;  the 
number  of  bushels  in  a  given  excavation  expressed  in  yards 
may  be  obtained  from  the  formula 

C=.0495x  Y 

14.  The  circumference  of  a  circle  may  be  obtained  from  the 
area  by  the  formula 

C=  3.5446  X  V2 

15.  The  area  of  the  surface  of  a  cylinder  may  be  expressed 
by  the  formula  A  =  (^C  X  L) -\-2a 

When  C  =  circumference 

L  =  length 
a  =  area  of  one  end 

16.  The  surface  of  a  sphere  may  be  expressed  by  the  formula 

S=D'x  3.1416 

17.  The  solidity  of  a  sphere  may  be  obtained  from  the 
formula 

S  =  D'x  .5236 

18.  The  side  of  an  inscribed  cube  of  a  sphere  may  be  ob- 
tained from  the  formula 

S  =  R  X  1.1547,     where  S  =  length  of  side, 

R  =  radius  of  sphere. 


304  VOCATIONAL  MATHEMATICS 

19.  The  solidity  or  contents  of  a  pyramid  may  be  expressed 
by  the  formula 

F 
S  =  Ax^,  where  A  =  area  of  base, 

F  =  height  of  pyramid. 

20.  The  length  of  an  arc  of  a  circle  may  be  obtained  from 
the  formula 

L  =  Nx  .017453  E,         where  L  =  length  of  arc, 

N=  number  of  degrees, 
a  =  radius  of  circle. 

21.  The  horse  power  of  a  single  leather  belt  may  be  deter- 
mined by  the  formula 

DRW 

HP  =      ^       ,       where  D  =  diameter  of  pulle}^  in  inches, 

W—  width  of  belt  in  inches, 

M  =  revolutions  per  minute, 

HP  =  horse  power  transmitted. 

22.  The  formula  for  finding  the  weight  of  an  iron  ball  may 
be  calculated  by  the  following : 

Tr=  2)3x0.1377 

23.  The  formula  for  finding  the  diameter  of  an  iron  ball 
when  the  weight  is  given  is 

D  =  1.936  Vl^       where  D  =  diameter  of  the  ball  in  inches, 
W=  weight  of  ball  in  pounds. 

24.  The  volume  of  a  sphere  when  the  circumference  of  a 
great  circle  is  known  may  be  determined  by  the  formula 

25.  The  diameter  of  a  circle  the  circumference  of  which  is 
known  may  be  found  by  the  formula 


FORMULAS  305 

26.  The  area  of  a  circle  the  circumference  of  which  is  known 
may  be  found  Jby  the  formula 

4  tr 

Coefficients  and  Similar  Terms 

When  a  quantity  may  be  separated  into  two  factors,  one  of 
these  is  called  the  coefficient  of  the  other ;  but  by  the  coefficient 
of  a  term  is  generally  meant  its  numerical  factor. 

Thus,  4  6  is  a  quantity  composed  of  two  factors  4  and  6 ;  4  is  a  coef- 
ficient of  h. 

Similar  terms  are  those  that  have  as  factors  the  same  letters 
with  the  same  exponents. 

Thus,  in  the  expression,  6  a,  4  6,  2  a,  6  a6,  5  a,  2  6.  6  a,  2  a,  5  a  are 
similar  terms  ;  4  6,  2  &  are  similar  terms  ;  5  ab  and  6  a  are  not  similar 
terms  because  they  do  not  have  the  same  letters  as  factors.  8  a&,  6  ah, 
1  aft,  8  ah  are  similar  terms.  They  may  be  united  or  added  by  simply 
adding  the  letters  to  the  numerical  sum,  17  aft. 

In  the  following,  8  ft,  6  ft,  3  aft,  4  a,  aft,  and  2  a,  8ft  and  5ft  are  similar 
terms  ;  3  aft  and  aft  are  similar  terms  ;  4  a  and  2  a  are  similar  terms  ;  8  ft, 
3  aft,  and  4  a  are  dissimilar  terms. 

In  addition  the  numerical  coefficients  are  algebraically  added ; 
in  subtraction  the  numerical  coefficients  are  algebraically  sub- 
tracted ;  in  multiplication  the  numerical  coefficients  are  alge- 
braically multiplied ;  in  division  the  numerial  coefficients  are 
algebraically  divided. 

EXAMPLES 

State  the  similar  terms  in  the  following  expressions : 

1.  5  a;,  8  OLc,  3  a;,  2  ax.  6.    15  ahc,  2  aJbc,  4  dbc^  2  aft, 

2.  8aftc^  7c,  2aft,  3c,  ^ah,    3aZ>. 

9«*c.  7.   Saj,  6a;,  13xy,  6x,  7y. 

3.  2pq,  5p,Sq,  2p,  3g,  ^pq.        a    7y,  2y,  2xy,  3y,  2xy, 

4.  4.V,  5y2,  2y,ir)2,52,  2.V2.  ir       .  „ 

«    1ft         r       K      A         o  ^    2ir,  5xr*, -,  irr»,  2xr. 

5.  lo  mn,  0  m,  5  7i,  4  mny  2  m.  J 


306  VOCATIONAL  MATHEMATICS 

Equations 

A  statement  that  two  quantities  are  equal  may  be  expressed 
mathematically  by  placing  one  quantity  on  the  left  and  the 
other  on  the  right  of  the  equality  sign  (=).  The  statement 
in  this  form  is  called  an  equation. 

The  quantity  on  the  left  hand  of  the  equation  is  called  the 
left-hand  member  and  the  quantity  on  the  right  hand  of  the 
equation  is  called  the  right-hand  member. 

An  equation  may  be  considered  as  a  balance.  If  a  balance 
is  in  equilibrium,  we  may  add  or  subtract  or  multiply  or  divide 
the  weight  on  each  side  of  the  balance  by  the  same  weight  and 
the  equilibrium  will  still  exist.  So  in  an  equation  we  may 
perform  the  following  operations  on  each  member  without 
changing  the  value  of  the  equation : 

We  may  add  an  equal  quantity  or  equal  quantities  to  ea/ih  memr 
her  of  the  equation. 

We  may  subtract  an  equal  quantity  or  equal  quantities  from 
each  member  of  the  equation. 

We  may  multiply  each  member  of  the  equation  by  the  same  or 
equal  quantities. 

We  may  divide  each  member  of  the  equation  by  the  same  or 
equal  quantities. 

We  may  extract  the  square  root  of  each  member  of  the  equation. 

We  may  raise  each  member  of  the  equation  to  the  same  poiver. 

The  expression,  A  =  ttR^  is  an  equation.     Why  ? 

If  we  desire  to  obtain  the  value  of  R  instead  of  A  we  may  do 
so  by  the  process  of  transformation  according  to  the  above 
rules.  To  obtain  the  value  of  R  means  that  a  series  of  opera- 
tions must  be  performed  on  the  equation  so  that  R  will  be  left 
on  one  side  of  the  equation. 

(1)  A  =  7rJ?2 

A 

(2)  —  =  R^     (Dividing  equation  (1)  by  the  coefficient  of  B^.) 

IT 

(^)     -\/~  =  -^     (Extracting  the  square  root  of  each  side  of  the  equation.) 


FORMULAS  307 

Methods  of  Representing  Operations 
Multiplication 

The  multiplication  sign  ( X )  is  used  in  most  eases.  It  should 
not  be  used  in  operations  where  the  letter  (x)  is  also  to  be  em- 
ployed. 

Another  method  is  as  follows  : 

2.3    a. 6     2  a. 36     4a;. 5a 

This  method  is  very  convenient,  especially  where  a  number 
of  small  terms  are  employed.  Keep  the  dot  above  the  line, 
otherwise  it  is  a  decimal  point. 

Where  parentheses,  etc.,  are  used,  multiplication  signs  may 
be  omitted.  For  instance,  (a  +  b)x{a  —  h)  and  (a  -\-  b'){a  —  b) 
are  identical ;  also,  2'(x  —  y)  and  2{x  —y). 

The  multiplication  sign  is  very  often  omitted  in  order  to 
simplify  work.  To  illustrate,  2  a  means  2  times  a ;  5  xyz  means 
5  •  X  •  y  •  z ;  x{a  —  b)  means  x  times  (a  —  b),  etc. 

A  number  written  to  the  right  of,  and  above,  another  (a:*)  is 
a  sign  indicating  the  special  kind  of  multiplication  known  as 
involution. 

In  multiplication  we  add  exponents  of  similar  terms. 

Thus,  «* .  a^  =  a:»+'  =  ar* 

(ibc  '  ab  •  aVj  =  a*bh 
The  multiplication  of  dissimilar  terms  may  be  indicated. 
Thus,  a-b  '  C'  x-y  'Z  =  abcxyz. 

Division 

The  division  sign  (-*-)  is  used  in  most  cases.  In  many 
cases,  however,  it  is  best  to  employ  a  horizontal  line  to  indicate 

division.     To  illustrate, means  the  same  as  (a  -\'  b)  -i- 

x-y 

(x  —  y)  in  simpler  form.     The  division  sign  is  never  omitted. 


308  VOCATIONAL  MATHEMATICS 

A  root  or  radical  sign  (V^>  y/^)  is  a  sign  indicating  the  special 
form  of  division  known  as  evolution. 

In  division,  we  subtract  exponents  of  similar  terms. 

Thus,  ar»H-aj'  =  -  =  a^-2  =  a; 


a^lfif^  ^  a'^bc^  =  ^^^^^  =  a^bc. 
a^bc^ 

The  division  of  dissimilar  terms  may  be  indicated. 

Thus,  (abc)  -i-  xyz  = 

xyz 

Substituting  and  Transposing 

A  formula  is  usually  written  in  the  form  of  an  equation. 
The  left-hand  member  contains  only  one  quantity,  which  is 
the  quantity  that  we  desire  to  find.  The  right-hand  member 
contains  the  letters  representing  the  quantity  and  numbers 
whose  values  we  are  given  either  directly  or  indirectly. 

To  find  the  value  of  the  formula  we  must  (1)  substitute  for 
every  letter  in  the  right-hand  member  its  exact  numerical 
value,  (2)  carry  out  the  various  operations  indicated,  remem- 
bering to  perform  all  the  operations  of  multiplication  and 
division  before  those  of  addition  and  subtraction,  (3)  if  there 
are  any  parentheses,  these  should  be  removed,  one  pair  at  a 
time,  inner  parentheses  first.  A  minus  sign  before  a  parenthesis 
means  that  when  the  parenthesis  is  removed,  all  the  signs  of 
the  terms  included  in  the  parenthesis  must  be  changed. 

Find  the  value  of  the  expression 

3d-\-b(2a-b-\-  18),  where  a  =  5,b  =  3. 

Substitute  the  value  of  each  letter.  Then  perform  all  addition  or 
subtraction  in  the  parentheses. 

3x6-1-  3(10  -  3  +  18) 
16-^.3(28-3) 
15  +  3(26) 
15  4-75  =  90 


FORMULAS  309 

EXAMPLES 
Find  the  value  of  the  following  expressions : 

1.  2  ^  X  (2  +  3  ^)  X  8,  when  J  =  10. 

2.  8  o  X  (6  —  2  a)  X  7,  when  a  =  7. 

3.  8  6  +  3  c  +  2  rt (a  -f  />  +  c)  -  8, when  a  =  9;  6  =  11 ;  c  =  13. 

4.  8(a;  +  y),  when  a?  =  9;  y  =  11. 

5.  13  {x-  y),  when  a;  =  27 ;  y  =  9. 

a   24y-|-8z(2  +  y)-3y,  when  jy  =  8;  z  =  ll. 

7.    Q{(^M-\-^N)-^2  0,  when  Jlf=4,  xV=5,  Q=6,  0  =  8. 

a   Find  the  value  of  X  in  the  formula  X  =  ^^^^^"^^^ 
when  J»f  =  11,  iV^=  9,  P  =  28. 

9.    .c  =  ^.^L±^,  when  n  =  5,  m  =  6,  P  =  8,  Q  =  7. 

10.  Find  the  value  of  T  in  the  equation 

(x  +  y)(x-y) 

11.  3a+4(6-2a  +  3c)-c,  when  a  =  4,  6  =  6,  c  =  2. 

12.  op  —  Sq(p  +  r  —  S)  —  qy  when  /)  =  5,  g  =  7,  r  =  9,  i5  =  11. 

13.  S^^t*-p'i-3(S-\-t  +  p)j  whenp  =  5,  S=Sy  t  =  9. 

14.  a*  -  &>+  c»,  when  a  =  9,  6  =  6,  c  =  4. 

15.  (a  +  6)  (a  +  6  -  c),  when  a  =  2,  6  =  3,  c  =  4. 

16.  (a*  -  b*)  (a*  +  6*),  when  a  =  8,  6  =  4. 

17.  (c-»  +  cP)  (c»  -  rf»),  c  =  9,  (/  =  5. 


18.  Va*  +  2  a6  +  6«,  when  a  =  7,  6  =  8. 

19.  ^c*-61,  when  c  =  5. 


310  VOCATIONAL  MATHEMATICS 

PROBLEMS 

Solve  the  following  problems  by  first  writing  the  formula 
from  the  rule  on  page  300,  and  then  substituting  for  the  answer. 

1.  How  many  electrical  horse  power  in  4389  watts  ? 

2.  How  many  kilowatts  in  2389  watts  ? 

3.  (a)  Give  the  number  of  watts  in  a  circuit  of  110  volts 
and  25  amperes. 

(6)  How  many  electrical  horse  power  ? 

4.  What  is  the  voltage  of  a  circuit  if  the  horse  power  is 
2740  watts  and  the  quantity  of  electricity  delivered  is  25 
amperes  ? 

5.  What  is  the  resistance  of  a  circuit  if  the  voltage  is  110 
and  the  quantity  of  electricity  is  25  amperes  ? 

6.  What  is  the  pressure  per  square  inch  of  water  87  feet 
high? 

7.  What  is  the  capacity  of  a  cylinder  with  a  base  of  16 
square  inches  and  6  inches  high  ?  (Capacity  in  gallons  is 
equal  to  cubical  contents  obtained  by  multiplying  base  by  the 
height  and  dividing  by  231  cubic  inches.) 

8.  What  is  the  length  of  a  30°  arc  of  a  circle  with  16" 
diameter? 

9.  What  is  the  area  of  a  sector  with  an  arc  of  40°  and  a 
diameter  of  18"? 

10.  What  is  the  weight  of  the  rim  of  a  flywheel  of  a  25  H.  P. 
engine  ? 

11.  What  is  the  area  of  a  cylinder  of  a  50  H.  P.  engine  with 
the  piston  making  120  ft.  per  minute  ? 

12.  What  is  the  H.  P.  of  a  2ff''  shaft  making  180  R.  P.  M.  ? 

13.  What  is  the  capacity  of  a  pail  14''  (diameter  of  top), 
11"  (diameter  of  bottom),  and  16"  in  height  ? 

14.  What  is  the  area  of  an  ellipse  with  the  greatest  length 
16"  and  the  greatest  breadth  10"  ? 


FORMULAS 


311 


Interpretation  of  Negative  Quantities 

The  quantity  or  number  —  12  lias  no  meaning  to  us  according 
to  our  knowledge  of  simple  arithmetic,  but  in  a  great  many 
problems  in  practical  work  the  minus  sign  before  a  number 
assists  us  in  understanding  the  different  solutions. 

To  illustrate : 


FaBRSNHKIT  TllF.BlIOllBTKR 


Cbntiorape  Tiibrmombtkr 


Boilinf? 
point  of 


Freezinjf 
point  of  ■ 
water 


212* 


Roilinsr 
point  uf 
water 


82* 


2.  a. 


Freezing 
point  uf  ' 
water 


100» 


On  the  Centigrade  scale  the  freezing  point  of  water  is  marked 
0°.  Below  the  freezing  point  of  water  on  the  Centigrade  scale 
all  readings  are  expressed  as  minus  readings. 

—  30°  C  means  thirty  degrees  below  the  freezing  point.  In 
other  words,  all  readings,  in  the  direction  below  zero  are 
expressed  as  — ,  and  all  readings  above  zero  are  called  -|-. 
Terms  are  quantities  connected  by  a  plus  or  minus  sign. 
Those  preceded  by  a  plus  sign  (when  no  sign  precedes  a  quan- 
tity plus  is  understood)  are  called  positive  quantities,  while 
those  connected  by  a  minus  sign  are  called  negative  quantities. 


312  VOCATIONAL  MATHEMATICS 

Let  us  try  some  problems  involving  negative  quantities. 
Find  the  corresponding  reading  on  the  Fahrenheit  scale  cor- 
responding to  — 18°  C. 

F  =  I  C  +  32° 
F=  |(- 18°)+32° 

Notice  that  a  minus  quantity  is  placed  in  parenthesis  when  it  is  to  be 
multiplied  by  another  quantity. 

F  =-  ip°  +  32"  =  -  32f°  +  32°  ;  F  =  -  |°. 

The  value  —  §°  is  explained  by  saying  it  is  §  of  a  degree  below  zero 
point  on  Fahrenheit  scale. 

Let  us  consider  another  problem.  Find  the  reading  on  the  Centi- 
grade scale  corresponding  to  —  40°  F. 

Substituting  in  the  formula,  we  have 

C  =  s  (_  40°  _  32C) ^  |(_  72)  =  _  40^ 

Since  subtracting  a  negative  number  is  equivalent  to  adding 
a  positive  number  of  the  same  value,  and  subtracting  a  posi- 
tive number  is  equivalent  to  adding  a  negative  number  of  the 
same  value,  the  rule  for  subtracting  may  be  expressed  as  fol- 
lows: Change  the  sign  of  the  subtrahend  and  proceed  as  in 
addition. 

For  example,  40  minus  —  28  equals  40  plus  28,  or  68. 

40  minus  +  28  equals  40  plus  —  28,  or  12. 

—  40  minus  4-  32  equals  —  40  plus  —  32  =  -  72. 

(Notice  that  a  positive  quantity  multiplied  by  a  negative  quantity  or 
a  negative  quantity  multiplied  by  a  positive  quantity  always  gives  a 
negative  product.  Two  positive  quantities  multiplied  together  will  give 
a  positive  product,  and  two  negative  quantities  multiplied  together  will 
give  a  positive  product.)     To  illustrate  : 

5  times  5  =  5  x  5  =  25 

5  times  — 5=r5x(—  5)  =  —  25 

(-5)  times  (-5)  =  + 25 

In  adding  positive  and  negative  quantities,  first  add  all  the 
positive  quantities  and  then  add  all   the  negative  quantities 


FORMULAS  313 

together.     Subtract  the  smaller  from  the  larger  and  prefix  the 
same  sign  before  the  remainder  as  is  before  the  larger  number. 

For  example,  add : 

2a,  5a,  -  Oa,  8a,  -2a 
2a  +  6a -I- 8a  =  16a;  -6a-2a=-8a 
T5a-8a  =  7a 

EXAMPLES 
Add  the  following  terms : 

1.  3Xf  —X,  7  Xf  AXj  —2x. 

2.  6y,  2y,  9y,  -7y. 

3.  9  aby  2  a6,  6  aft,  —  4  aft,  7  a6,  —  5  ab. 

Multiplication  of  Algebraic  Expressions 

Each  term  of  an  algebraic  expression  is  composed  of  one  or 
more  factors,  as,  for  example,  2  ab  contains  the  factors  2,  a,  and 
b.  The  factors  of  a  term  have,  either  expressed  or  understood, 
a  small  letter  or  number  in  the  upper  right-hand  corner,  which 
states  how  many  times  the  quantity  is  to  be  used  as  a  factor. 
For  instance,  ab\  The  factor  a  has  the  exponent  1  understood 
and  the  factor  b  has  the  exponent  2  expressed,  meaning  that  a 
is  to  be  used  once  and  b  twice  as  a  factor,  ab'^  means,  then, 
a  X  6  X  6.  The  rule  of  algebraic  multiplication  by  terms  is  as 
follows:  Add  the  exponents  of  all  like  letters  in  the  terms 
multiplied  and  use  the  result  as  exponent  of  that  letter  in  the 
product.  Multiplication  of  unlike  letters  may  be  expressed 
by  placing  the  letters  side  by  side  in  the  product. 

For  example  :  2  a6  x  .3  6^  =  0  a6« 

4ax86  =  12a6 

Algebraic  or  literal  expressions  of  more  than  one  term  are 
multiplied  in  the  following  way :  begin  with  the  first  term  to 
the  left  in  the  multiplier  and  multiply  every  term  in  the  multi- 
plicand, placing  the  partial  products  underneath  the  line.    Then 


314  VOCATIONAL  MATHEMATICS 

repeat  the  same  operation,  using  the  second  term  in  the  multi- 
plier. Place  similar  products  of  the  same  factors  and  degree 
(same  exponents)  in  same  column.     Add  the  partial  products. 

Thus,  a  -\-  b  multiphed  by  a  —  5. 

a  +  h 
a  —  h 


a^+  ab-  b-^ 
-ah 


a^  -  b-^ 

Notice  the  product  of  the  sum  and  difference  of  the  quantities  is  equal 
to  the  difference  of  their  squares. 

EXAMPLES 

1.  Multiply  a-\-hhy  a-\-h. 

State  what  the  square  of  the  sum  of  the  quantities  equals. 

2.  Multiply  X  —  y  hy  X  —  y. 

State  what  the  square  of  the  difference  of  the  quantities  equals. 

3.  Multiply  (j)  +  q)(p  —  q).  7.    Multiply  (x  —  y)(x  —  y). 

4.  Multiply  (p  -f  q)(p  +  q).  8.    (x  -\-yy=? 

5.  Multiply  (r  +  s){r  -  s).  9.    {x  -  yf  =  ? 

6.  Multiply  (a  ±  6)(a  ±  6).  10.    (x  +  y){x-y)  =  ? 


LOGARITHMS 

Thk  logarithm  of  a  iuiiuImt  to  the  base  10  is  defined  as  the  povrer 
to  Nvhicli  10  must  be  raiseil  in  order  to  equal  the  number. 

Thus  the  logarithm  (or  log,  as  it  is  more  generally  written)  of  10 
Ls  1,  for  the  first  power  of  10  is  10.  The  log  of  100  is  2,  for  the  second 
power  of  10  is  100.  Hence  the  log  of  a  number  between  10  and 
100  is  a  number  between  1  and  2,  and  is,  therefore,  1  plus  a  decimal. 
The  whole  number  1  is  called  the  characteristic,  and  the  decimal  part 
is  called  the  m<intissa.  In  like  manner,  the  log  of  a  number  between 
100  and  1000  is  2  plus  a  decimal,  for  10*  =  100  and  10«  =  1000.  The 
log  of  a  number  between  1  and  10  is  0  plus  a  decimal. 

The  Characteristic.  —  The  characteristic  of  the  log  of  a  number 
l)etween  1  and  10  is  0;  that  is,  the  characteristic  of  the  log  of  a 
number  that  has  one  digit  is  0.  The  characteristic  of  the  log  of  a 
number  between  10  and  100  is  1.  The  characteristic  of  the  log  of 
a  number  between  100  and  1000  is  2.  In  each  case,  the  character- 
istic is  seen  to  be  one  less  than  the  number  of  digits  at  the  left  of 
the  decimal  point.     Hence  the  rule : 

Rule  I. —  The  characteristic  of  the  logarithm  of  a  number  is  one  less 
than  the  number  of  digits  (or  the  number  of  figures)  in  the  numbers  that 
are  at  the  lefi  of  the  decimal  point. 

If,  however,  there  are  no  figures  at  the  left  of  the  decimal  point; 
that  is,  if  the  number  is  less  than  1,  we  must  use  the  following  rule: 

Rule  II. —  The  characteristic  of  a  number  less  than  one  is  (minus) 
one  more  than  the  number  of  zeros  at  the  right  (f  the  decimal  point  before 
the  first  significant  figure. 

Tims  the  characteristic  of  the  log  of  .41  is  —  1 ;  of  0.0302  is  —  2; 
of  0.0032  is  -  3,  etc.  These  are  usually  written  9  -^  10;  8-10; 
7-  10. 

The  Mantissa.  —  Consider  the  three  numV>ers  7.21,  72.1,  and  721. 
721.  =  10  X  72.1  =  10*  X  7.21.  Therefore,  log  721  =  log  10  +  log  72.1 
=  log  10*  +  log  7.21.  (In  order  to  multiply  the  numbers,  we  may 
add  the  logarithms  of  these  numbers.)  Therefore,  log  721  =  1  + 
log  72.1  =  2  +  log  7.21,  (log  10  =  1 ;  log  10*  =  2).  Therefore,  we  see 
that  the  logarithms  of  numbers  made  up  of  the  same  sequence  of 

315 


316  VOCATIONAL  MATHEMATICS 

digits,  but  differing  in  the  position  of  the  decimal  point,  differ  only 
in  the  characteristic,  the  mantissce  remaining  the  same. 

For  this  reason,  in  making  tables  of  logaritlims,  only  the  mantissse 
are  given,  the  characteristics  being  added  according  to  the  rules 
above.  The  tables  we  are  to  use  are  made  for  three  significant  digits 
in  the  number  and  four  digits  in  the  mantissa  of  the  log. 

To  Find  the  Logarithm  of  a  Number 

A.  When  the  number  has  three  significant  figures. 

The  first  two  figures  are  found  in  the  column  headed  No. 
(pages  318-319),  and  the  third  figure  in  the  top  row.  The  mantissa 
is  stated  in  the  column  and  row  so  determined. 

To  illustrate,  find  the  log  of  62.8.  In  the  62  row  and  in  the  col- 
umn under  8  is  the  mantissa,  7980.  By  Rule  I  we  find  the  charac- 
teristic to  be  1 ;  therefore,  the  log  of  62.8  is  1.7980. 

Find  the  log  of  .00709.  The  significant  figures  are  709;  in  the 
70  row  and  in  the  column  under  9,  we  find  the  mantissa  8506.  By 
Rule  II  the  characteristic  is  —  3.  Therefore  the  log  of  .00709  is 
—  3.8506.  To  prevent  confusion  in  the  operations  of  arithmetic  —  3  is 
usually  written  3,  so  the  log  is  stated  to  be  3.8506,  or  7.8506  —  10. 

EXAMPLES 
Find  log  of  117  ;  of  1280  ;  of  16.5  ;  of  2.09  ;  of  .721 ;  of  .0121. 

B.  When  the  number  has  four  or  more  significant  figures. 

Ex.  1.  —  Find  the  log  of  7987.  The  tables  give  only  the  mantissa  of 
numbers  of  three  figures,  so  the  mantissa  of  7987  cannot  be  found  in 
the  table  ;  7987,  however,  lies  between  7980  and  7990,  hence  its  log  is 
between  the  log  of  7980  and  the  log  of  7990.  We  find  the  mantissa 
7980  to  be  9020,  and  the  mantissa  of  7990  to  be  9025.  Now  the 
difference  between  these  mantissas  is  5,  while  the  difference  between 
the  two  numbers,  7980  and  7990,  is  10.  But  the  difference  between 
7980  and  7987  is  ^jj  of  the  difference  between  7980  and  7990,  and  the 
mantissa  of  7987  will  be  /^  of  the  difference  between  the  mantissa  of 
7980  and  the  mantissa  of  7990.  Therefore  it  will  equal  9020  (the 
mantissa  of  7980)  plus  yV  of  5  (the  difference  between  the  mantissse 
of  7980  and  7990)  :  5  x  .7  =  3.5,  hence  the  mantissa  of  7987  is  9024. 
In  such  reckoning  all  decimal  parts  less  than  .5  are  counted  as  0,  and 
all  decimal  parts  greater  than  .5  are  counted  as  1 ;  .5  is  counted  as 
either  1  or  0.     Log  7987  is  .3^024. 


LOGARITHMS  317 

Ex.  2.  —  Find  the  log  of  12.564. 

First  find  the  mantissa.  In  looking  for  the  mantissa  the  decimal 
{H)int  need  not  be  considered. 

Solution.  — 12554  lies  between  12500  and  12600. 
Mantissa  of  12600  is  1004. 
Mantissa  of  12.500  is  0969. 

35  is  the  difference  between  the  mantissa;. 
12600  -  12500  =  100  =  difference  between  numbers. 
12554  —  12500  =  54  =  difference  between  original  numbers. 
The  multiplier  is  f*^  =  .54. 

35  X  .54  =  18.90  or  19. 
0969  -f  19  =  0988. 
log  12.564  is  1.0988.     Ans. 

EXAMPLES 

Find  log  of  17.89 ;  of  2172 ;  of  652.12 ;  of  4213  ;  of  3342000. 

C   To  find  the  number  corresponding  to  a  given  logarithm. 

Ex.  1.  —  What  number  has  the  log  1.6085? 

The  mantissa  6085  is  found  in  the  40  row  and  in  the  column 
under  6.  Tiie  number  corresponding  to  mantissa  6085  is  therefore 
406.  The  characteristic  1  states  that  there  are  two  digits  to  the  left 
of  the  decimal  point.     The  number  is  therefore  40.6. 

Ex.  2.—  What  number  has  the  log  5.8716? 

The  mantissa  8716  is  in  the  74  row  and  in  the  column  under  4. 
The  characteristic  5  states  that  there  are  six  digits  to  left  of  decimal 
point.     The  number  is  therefore  744000. 

Ex.  3.  —  What  number  has  the  log ,2.6538? 

The  mantissa  6538  is  not  found  in  the  tables,  but  lies  between 
6532  and  6542,  hence  the  number  (not  considering  the  decimal  point) 
lies  between  4.50  and  451.  The  difference  between  the  mantissa;  of 
4.50  and  451  is  10;  the  difference  between  the  mantissa  of  450  and  of 
the  number  to  be  found  is  6.  The  difference  between  450  and  451  is  1. 
1  X  .6  =  .6.  The  number  corresponding  to  mantissa  65.38  is  4506, 
hence  the  numl)er  corre.sponding  to  log  2.6538  is  4.50.6. 

EXAMPLES 

Find  number  corresponding  to  log  1.5481;  to  log  0.6681;  to  log 
1.9559;  to  log  2.9324. 


31S 


LOGARITHMS  OF  NUMBERS 


No. 

0 

1 

2 

3 

4 

0 

0 

7 

8 

9 

10 

OOCX) 

0043 

0086 

0128 

0170 

0212 

0253 

0294 

0334 

0374 

IX 

0414 

0453 

0492 

0531 

0569 

0607 

0645 

0682 

0719 

0755 

12 

0792 

0828 

0864 

0899 

0934 

0969 

1004 

1038 

1072 

1 106 

13 

1 139 

1173 

1206 

1239 

1271 

1303 

1335 

■^3^1 

1399 

1430 

14 

1 461 

1492 

1523 

1553 

1584 

1614 

1644 

1673 

1703 

1732 

15 

1 761 

1790 

1818 

1847 

1875 

1903 

1931 

1959 

1987 

2014 

16 

2041 

2068 

2095 

2122 

2148 

2175 

2201 

2227 

2253 

2279 

17 

2304 

2330 

2355 

2380 

2405 

2430 

2455 

2480 

2504 

2529 

18 

2553 

2577 

2601 

2625 

2648 

2672 

2695 

2718 

2742 

2765 

19 

2788 

2810 

2833 

2856 

2878 

2900 

2923 

2945 

2967 

2989 

20 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

21 

3222 

3243 

3263 

3284 

3304 

3324 

3345 

3365 

3385 

3404 

22 

3424 

3444 

3464 

3483 

3502 

3522 

3541 

3560 

3579 

3598 

23 

3617 

3636 

3655 

3674 

3692 

3711 

3729 

3747 

3766 

3784 

24 

3802 

3820 

3838 

3856 

3874 

3892 

3909 

3927 

3945 

3962 

25 

3979 

3997 

4014 

4031 

4048 

4065 

4082 

4099 

4116 

4133 

26 

4150 

4166 

4183 

4200 

4216 

4232 

4249 

4265 

4281 

4298 

27 

43H 

4330 

4346 

4362 

4378 

4393 

4409 

4425 

4440 

4456 

28 

4472 

4487 

4502 

4518 

4533 

4548 

4564 

4579 

4594 

4609 

29 

4624 

4639 

4654 

4669 

4683 

4698 

4713 

4728 

4742 

4757 

30 

4771 

4786 

48CX) 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

31 

4914 

4928 

4942 

4955 

4969 

4983 

4997 

501 1 

5024 

5038 

32 

5051 

5065 

5079 

5092 

5105 

5119 

5132 

5145 

5159 

5172 

33 

5185 

5198 

5211 

5224 

5237 

5250 

5263 

5276 

5289 

5302 

34 

5315 

5328 

5340 

5353 

5366 

5378 

5391 

5403 

5416 

5428 

35 

5441 

5453 

5465 

5478 

5490 

5502 

5514 

5527 

5539 

5551 

36 

55^3 

5575 

5587 

5599 

561 1 

5623 

5635 

5647 

5658 

5670 

37 

5682 

5694 

5705 

5717 

5729 

5740 

5752 

5763 

5786 

38 

5798 

5809 

5821 

5832 

5843 

5855 

^Hl 

5888 

5899 

39 

59" 

5922 

5933 

5944 

5955 

5966 

5977 

5988 

5999 

6010 

40 

6021 

6031 

6042 

6053 

6064 

6075 

6085 

6096 

6107 

6117 

41 

6128 

6138 

6149 

6160 

6170 

6180 

6191 

6201 

6212 

6222 

42 

6232 

6243 

6253 

6263 

6274 

6284 

6294 

6304 

6314 

6325 

43 

6335 

6345 

6355 

6365 

6375 

6385 

6395 

6405 

6415 

6425 

44 

6435 

6444 

6454 

6464 

6474 

6484 

6493 

6503 

6513 

6522 

45 

6532 

6542 

6551 

6561 

6571 

6580 

6590 

6599 

6609 

6618 

46 

6628 

6637 

6646 

6656 

6665 

6675 

6684 

6693 

6702 

6712 

47 

6721 

6730 

6739 

6749 

6758 

6767 

6776 

6785 

6794 

6803 

48 

6812 

6821 

6830 

6839 

6848 

6857 

6866 

6875 

6884 

6893 

49 

6902 

691 1 

6920 

6928 

6937 

6946 

6955 

6964 

6972 

6981 

50 

6990 

6998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7067 

51 

7076 

7084 

7093 

7IOI 

7110 

7118 

7126 

7135 

7143 

7152 

52 

7160 

7168 

7177 

7185 

7193 

7202 

7210 

7218 

7226 

7235 

53 

7243 

7251 

7259 

7267 

7275 

7284 

7292 

7300 

7308 

7316 

54 

7324 

7332 

7340 

7348 
3 

7356 

7364 

7372 

7380 

7388 

7396 

No. 

0 

1 

2 

4 

5 

6 

7 

8 

9 

LOGARITHMS  OF   NUMBERS 


310 


Ko. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

55 

7404 

7412 

7419 

7427 

7435 

7443 

745J 

7459 

7466 

7474 

56 

74S2 

7490 

7497 

7505 

75U 

7520 

7528 

7536 

7543 

755» 

57 

7559 

7566 

7574 

7582 

7589 

7597 

7604 

7612 

7619 

7627 

58 

7634 

7642 

7649 

7657 

7664 

7672 

7679 

7686 

7694 

7701 

59 

7709 

7716 

7723 

7731 

7738 

7745 

7752 

7760 

7767 

7774 

60 

7782 

7789 

7796 

7803 

7810 

7818 

7825 

7832 

7839 

7846 

6x 

7853 

7S60 

7808 

7875 

78S2 

7889 

7896 

7903 

7910 

7917 

6a 

63 

7924 
7993 

?^ 

7938 
8007 

7945 
8014 

7952 
8021 

^51 

7966 
8035 

7973 
8041 

c 

79S7 
8055 

64 

8062 

8069 

8075 

8082 

8089 

8096 

8102 

8109 

8II6 

8122 

65 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

66 

8195 

8202 

8209 

821S 

8222 

8228 

8235 

8241 

8248 

8254 

67 

8261 

8267 

8274 

8280 

8::S7 

8293 

8299 

8306 

8312 

8319 

C8 

8325 

8331 

8338 

8344 

8351 

8357 

8363 

8370 

8376 

8382 

69 

8388 

8395 

8401 

8407 

8414 

8420 

8426 

8432 

8439 

8445 

70 

8451 

8457 

8463 

8470 

8476 

8482 

8488 

8494 

8500 

8506 

71 

8513 

8519 

8525 

8531 

8537 

8543 

8549 

8555 

8561 

8567 

7a 

8573 

8579 

8585 

8591 

8597 

8603 

8609 

8615 

8621 

8627 

73 

8633 

8639 

8645 

8651 

8657 

8663 

8669 

8675 

8681 

8686 

74 

8692 

8698 

8704 

8710 

8716 

8722 

8727 

8733 

8739 

8745 

75 

8751 

8756 

8762 

8768 

8774 

8779 

8785 

8791 

8797 

8802 

76^ 

8808 

8814 

8820 

8825 

8831 

8837 

8842 

8848 

8854 

8859 

77 

8865 

8871 

8876 

8882 

8887 

8893 

8899 

8904 

8910 

8915 

78 

8921 

8927 

8932 

8938 

8943 

8949 

8954 

8960 

8965 

8971 

79 

8976 

8982 

8987 

8993 

8998 

9004 

9009 

9015 

9020 

9025 

80 

9031 

9036 

9042 

9047 

9053 

9058 

9063 

9069 

9074 

9079 

8x 

9085 
9138 

9090 

9096 

9101 

9106 

9112 

9117 

9122 

9128 

9»33 

8a 

9143 

9149 

9154 

9159 

9165 

9170 

9175 

9180 

9186 

83 

9191 

9196 

9201 

9206 

9212 

9217 

9222 

9227 

9232 

9238 

84 

9243 

9248 

9253 

9258 

9263 

9269 

9274 

9279 

9284 

9289 

85 

9294 

9299 

9304 

9309 

9315 

9320 

9325 

9330 

9335 

9340 

86 

9345 

9350 

9355 

9360 

9365 

9370 

9375 

9380 

9385 

9390 

87 

9395 

9400 

9405 

9410 

9415 

9420 

9425 

9430 

9435 

9440 

88 

9445 

9450 

9455 

9460 

9465 

9469 

9474 

9479 

9484 

9489 

89 

9494 

9499 

9504 

9509 

9513 

9518 

9523 

9528 

9533 

9538 

90 

9542 

9547 

9552 

9557 

9562 

9566 

9571 

9576 

9581 

9586 

91 

9590 

9595 

9600 

9605 

9609 

9614 

9619 

9624 

9628 

9633 

ga 

9638 

9643 

9647 

9652 

9657 

9661 

9666 

9671 

9675 

9680 

93 

968s 

9689 

9694 

9699 

9703 

9708 

9713 

9717 

9722 

9727 

94 

9731 

9736 

9741 

9745 

9750 

9754 

9759 

9763 

9768 

9773 

95 

9777 

9782 

9786 

9791 

9795 

9800 

9805 

9809 

9814 

9818 

96 

9823 

9827 

9832 

9836 

9841 

9845 

9850 

9854 

9859 

9863 

S7 

9868 

9872 

9877 

9881 

9886 

9890 

9894 

9899 

9903 

9908 

98 

9912 

9917 

9921 

9926 

9930 

9934 

9939 

9943 

9948 

9952 

99 

9956 

9961 

9965 

9969 

9974 

9978 

9983 

9987 

9991 

9996 
9 

Ho. 

0    1  1  2  !  3 

4 

5 

6 

7 

8 

TRIGONOMETRY 

There  are  many  problems  in  the  shop  that  involve  a  knowledge  of 
angles  and  the  relation  between  the  parts  of  triangles  —  angles  and 
sides.  Such  problems  include  laying  out  angles,  depth  of  screw 
threads,  diagonal  distance  across  bolts,  etc. 

Trigonometry  is  the  subject  which  deals  with  the  properties  and 
measurement  of  angles  and  sides  of  triangles,  and  is  spoken  of  in  the 
machine  shop  as  simply  "  trig." 

Trigonometric  Functions  (Ratios).  —  Since  the  sum  of  all  the  angles 
of  a  triangle  equals  180^,  and  since  a  right  triangle  is  composed  of  a 
right  angle  and  two  acute  angles,  it  follows  that,  if  we  know  one 
acute  angle,  we  may  obtain  the  other  by  subtracting  it  from  90°. 
The  angle  found  by  subtracting  a  definite  angle  from  90°  is  called 
the  complement  of  the  given  angle. 

The  complement  of  45°  is  45°,  for  90°  -  45°  =  45°. 

The  complement  of  30°  is  60°,  for  90°  -  30°  =  60°. 

The  supplement  of  an  angle  is  the  difference  between  it  and  180°. 
The  supplement  of  60°  is  120°,  for  180°  -  60°  =  120°. 

Examine  a  right  angled  triangle  very  carefully.  Notice  the  position 
of  the  two  acute  angles  and  of  the  right  angle.  The  longest  side  of 
the  triangle,  called  the  hypotenuse,  is  op- 
posite the  right  angle.  The  sides  of  the 
triangle  called  the  legs,  are  the  two 
smallest  sides.  These  sides  are  perpen- 
dicular to  each  other,  and  the  longer  side 
is  opposite  the  greater  angle. 

Draw  a  right    angle   with    sides    re-    .^ 

spectively  3  and  4  inches  long.  _  .  _ 

„,         ,         .1      ,         ,  ,  Right  Angled  Triangle 

Then  draw  the  hypotenuse  and  meas- 
ure the  length  of  it,  which  will  be  5  inches. 

Divide  the  length  of  the  side  opposite   Z  A  hy  the  hypotenuse, 

3  ^  5  =  .6. 

Divide  the  length  of  the  side  adjacent  to  Z  A  by  the  hypotenuse, 

4  -f-  5  =  .8. 

Divide  the  length  of  the  side  opposite  Z  A  hy  the  adjacent  side, 
3  -i-  4  =  .75. 

320 


TRIGONOMETRY  321 

These  values  are  called  ratios.  The  ratio  of  the  length  of  the  side 
opposite  Z  A  to  the  length  of  the  hypotenuse  is  called  the  sine  oi  ^  A. 

The  ratio  of  the  adjacent  Z  ^  to  the  length  of  hypotenuse  is 
called  the  cosine  of  Z  A. 

The  ratio  of  the  length  of  side  opposite  Z  A  to  the  length  of 
the  adjacent  leg  is  called  the  tangent  oi  Z  A. 

The  ratio  of  the  length  of  the  adjacent  side  of  Z  ^4  to  the  opposite 
side  is  called  the  cotangent. 

The  ratio  of  the  length  of  the  hypotenuse  to  the  adjacent  side  of 
Z  A  IB  called  the  secant. 

The  ratio  of  the  length  of  the  hypotenuse  to  the  opposite  side  of 
Z  A  is  called  the  cosecant. 

These  six  ratios  or  constants  represent  a  definite  relation  between 
the  sides  and  angles  of  right  triangles. 

Since  all  triangles  may  be  divided  into  right  angled  triangles,  by 
dropping  a  perpendicular  from  one  of  the  vertices  of  the  triangle,  we 
might  say  these  relations  apply  to  all  triangles.  The  same  relations 
apply  to  all  figures,  since  any  figure  may  be  divided  into  triangles, 
and  then  into  right  angled  triangles. 

These  definite  relations  between  the  sides  and  angles  are  called 
the  functions  of  the  angles,  and  are  i*eally  constants  representing  the 
fixed  proportions  between  the  sides  and  angles  of  a  triangle.  The 
exact  values  of  these  functions,  or  the  proportion  of  the  various  parts 
of  the  triangles,  have  been  figured  out  for  every  degree  that  will  be 
used  in  daily  practice.  They  are  given  in  the  tables  for  sine,  cosine, 
and  tangent  at  the  end  of  this  chapter. 

By  means  of  a  protractor  construct  an  angle  of  45°,  Z  CAB.     At 
a  point  on  ^  C  1"  from  A^  erect  a  perpendicular, 
CB.     Connecting  parts  A  and  B  by  the  line  AB, 
we  have  a  rt.  A  with  BC  =  AC,  because  Z  A  — 
45°  and  Z  B  =  45°. 

Measure  accurately  the  length  of  all  the  sides 
of  the  triangle. 

Divide  the  length  of  side  opposite  Z  A  by  the 

hypotenuse,  — '-  —  .7. 
^^  AB 

AC 

Divide  the  length  of  side  adjacent  to  ^A  by  the  hypotenuse,  —  =  .7. 

AB 


322  VOCATIONAL  MATHEMATICS 

Divide  the  length  of  side  opposite  Z  A  by  the  side  adjacent  to  ZA, 

AC 

Divide  the  length  of  the  adjacent  side  by  the  side  opposite  ZA, 

BC 

Divide  the  length  of  the  hypotenuse  by  the  side  adjacent  to  ZA, 

i^  =  1.4. 
AC 

Divide  the  length  of  the  hypotenuse  by  the  side  opposite  ZA, 

^^1.4. 
BC 

If  the  work  is  carried  out  accurately,  the  above  values  should  be 
obtained. 

Find  the  value  of  sine,  cosine,  and  tangent  when  angles  of  25^,  30°, 
60°,  75°,  are  constructed. 

The  reciprocal  of  the  tangent  of  an  angle  {Z  A)  is  called  the  co- 
tangent. 

The  reciprocal  of  the  sine  of  an  angle  (^Z  A)  is  called  the  cosecant. 

The  reciprocal  of  the  cosine  of  an  angle  {ZA)  is  called  the  secant. 

If  the  angles  of  the  triangles  are  always  marked  with  capital  letters, 
and  the  length  of  the  sides  opposite  the  angles  are  marked  with  small 
letters  as  shown  below,  it  will  be  possible  to  abbreviate  functions  of  the 
angle  in  terms  of  the  sides. 

Trigonometrical  Formulas,  etc. 

Geometrical  Solution  of  Right 
Angled  Triangles. 

c  =  Va^  +  62 
h  =  Vc^  -  a2 


a  =  Vc2  -  62 

.      a     opposite  side  „  j 

sin  A=-  =  -i-s- cos  A 

c       hypotenuse  c       hypotenuse 

A  _a_  opposite  side  +  j  _  ^  _  adjacent  side 

b     adjacent  side  a     opposite  side 

.      c       hypotenuse  ^^c««  a      c       hypotenuse 

sec  ^  =  -  =  --4-t^ ^^  cosec  A  =  -=  — •^^^-^~ — ^^ 

b     adjacent  side  a     opposite  side 


THIGONOMETHY  323 

Practical  Value  of  a  Trigonometric  Ratio 

Since  each  function  of  an  angle  may  be  expressed  as  the  ratio  of  two 
sides,  it  follows  that  if  we  know  the  size  of  an  angle  and  the  length  of 
one  side  of  a  right  triangle,  we  may  determine  the  other  sides  by  sub- 
stituting in  the  equation. 

Ex.  If  one  angle  of  a  right  angled  triangle  is  30^,  and  the  adjacent 
side  is  25  inches,  what  is  the  length  of  the  other  leg  ? 

D 

Since  we  desire  the  other  leg  we  should  ust 
(Formula  involving  two  legs) 


Tangent  ^A=^ 


25 


Ix>oking  up  tables  for  tangent  .1,  we  find  it  to  be  .5773.") ;  substitut- 
ing, we  have 


.577  =  —  a  =  14.4 

25 


EXAMPLES 

1.  The  simplest  application  of  trigonometry  in  the  shop  is  the 
problem  for  finding  the  depth  of  a  V  screw  thread.  Since  the  angle  at 
the  depth  of  the  screw  thread  is  60°  and  the  sides  are  equal,  an  equi- 
lateral triangle  is  formed.  Forming  right  angled  triangles  by  drop- 
ping a  perpendicular  from  the  angle  of  the  thread,  we  divide  the  angle 
of  00°  into  two  angles  of  30'^  each.  If  we  consider  the  pitch  1  inch, 
then  the  sides  are  each  an  inch  and  the  perpendicular  divides  the  base 
into  two  parts  of  \  inch  each. 

The  perpendicular,  or  depth  of  the  thread,  is  the  cosine  of  the 
angle  of  30°.  Looking  in  the  table  for  30°,  then  across  until  we 
come  to  the  column  headed  cosine,  we  find  .8.  This  gives  us  the 
depth  directly. 

2.  Find  the  side  of  the  thread  if  the  depth  is  1  inch. 

3.  What  is  the  depth  of  a  V  thread  with  8  threads  to  the  inch  ? 

4.  What  is  the  distance  across  the  corners  of  a  square  bar  3^"  on 
each  side  ? 


324  VOCATIONAL  MATHEMATICS 

6.  What  size  circular  stock  will  be  required  to  mill  a  square  nut 
lj\"  on  a  side? 

6.  What  is  the  distance  across  the  corners  of  a  bar  2^"  by  |"? 
What  angle  does  the  diagonal  make  with  the  base? 

Natural  Trigonometric  Functions 

The  tables  on  pages  326-329  give  the  numerical  values  of  the 
trigonometric  functions  of  angles  between  0°  and  90°  with  inter- 
vals of  10  minutes.  For  angles  from  0°  to  and  including  44°,  read 
from  the  top  of  the  table  downwards;  for  angles  from  45°  to  and  in- 
cluding 89°,  read  from  bottom  of  table  upwards.  The  degrees  and 
minutes  are  found  in  the  column  marked  A  or  Angle,  and  the  value  of 
the  function  in  the  column  marked  by  the  name  of  the  function.  For 
instance,  the  value  of  the  sine  of  15°  40'  is  found,  in  the  column 
marked  sine  and  in  the  row  40  under  15,  to  be  .2700.  The  value  of  the 
cotangent  of  63°  10'  is  found  (reading  from  the  bottom  of  the  table 
upwards,  since  the  angle  is  between  45°  and  89°),  in  the  column 
marked  cotangent  (at  bottom  of  table)  and  in  the  row  marked  10 
above  63,  to  be  .5059.  The  values  of  the  sine,  tangent,  and  secant  in- 
crease as  the  angle  increases,  while  the  values  of  the  cosine,  cotangent, 
and  cosecant  decrease  as  the  angle  increases,  so  that  for  the  former 
functions  the  correction  must  be  subtracted  from  the  value  given  in 
the  table. 

I.  To  find  the  value  of  a  function  of  an  angle  which  is  not  given  in 
the  table. 

Ex.  1.     Find  the  value  of  sine  35°  21'. 

Angle  35  is  given  in  table,  but  21'  is  not.  21',  however,  lies  between 
20'  and  30',  hence  the  value  of  sine  of  35°  21'  will  lie  between  the  value 
of  sine  of  35^  20'  and  the  value  of  sine  of  35°  30'.  The  value  of  sine  of 
35°  20'  is  given  in  table  as  .5783,  and  the  value  of  sine  of  35°  30'  is 
.5807 ;  the  difference  between  these  two  values  is  .0024.  The  tabular 
difference  in  the  angles  (that  is,  the  angular  difference  between  35°  20' 
and  35°  30')  is  10',  \\4iile  the  angular  difference  between  the  smaller 
angle  given  in  the  table  (35°  20')  and  the  angle  of  which  we  are  trying 
to  find  the  value  (35°  21')  is  1'.  The  correction  which  we  will  have  to 
add  to  .5783,  the  value  of  sine  of  35°  20',  is  j\  x  .0024  =  .00024.  Hence 
the  value  of  sine  of  35°  21'  is  .5783  -1-  .00024  =  .57854. 


TRIGONOMETRY  325 

Ex.  2.    Find  the  value  of  cotangent  82°  11'. 

82*'  41'  lies  between  82°  40'  and  82°  50 
cotan82°40'  =  .12869 
cotan  82°  oO'  =  .12574 
Difference  =  .00295 

The  tabular  difference  l>etween  angles  82°  40'  and  82°  50'  is  10'.  The 
difference  between  the  smaller  angle  and  given  angle  (82°  41'  —  82°  40') 
is  1'.  The  correction  is  therefore  y^^  x  .00295  or  .000295,  but  since  the 
last  figure  is  5  we  called  it  .00300.  This  correction  must  be  subtracted 
from  the  value  of  cotan  82°  40'  since,  as  above,  the  value  of  the  cotan 
decreases  as  the  angle  increases. 

cotan  82°  41'  =  .12869  -  .00.3  =  .12569 

EXAMPLES 

Find  value  of  tan  45°  19' ;  cosine  32°  8' ;  cotan  78°  51' ;  sine  62°  37'. 
II.   To  find  the  value  of  an  angle,  when  the  value  of  a  function  is 
given. 

Ex.  1.     Find  the  angle  whose  cotan  is  .6873. 

In  the  column  marked  cotan  look  for  the  numbers  .6873.  Since  it 
is  found  in  the  column  marked  cotan  at  the  bottom  of  the  table,  the 
degrees  and  minutes  will  be  found  at  the  right-hand  side  of  the  table, 
reading  from  the  bottom  upwards.  It  is  in  the  row  marked  30  above 
5.").     Hence,  angle  55°  30'  has  the  cotan  .6873. 

Ex.  2.     Find  the  angle  whose  cosine  is  .9387. 

This  number  is  found  in  the  column  marked  cosine  at  the  top  of  the 
table,  therefore  the  degrees  and  minutes  will  be  found  on  the  left  of 
the  table  reading  from  top  downwards.  It  is  in  the  row  marked  10 
under  20.     Hence,  the  angle  20"^  10'  has  the  cosine  .9387. 

EXAMPLES 

Find  the  angle  that  has  the  sine  .7642;  cotan  1..5607;  cosine  .4746; 
tangent  1.5108. 


326 


NATURAL  SINES  AND   COSINES 


A. 

Sin. 

Cos. 

A. 

Sin. 

Cos. 

|A. 

Sin. 

Cos. 

0° 

lO' 

20' 
30' 
40' 
50' 

1° 
10' 

20'- 

50' 
2° 
10' 
20' 
30' 
40; 
50' 
3° 
10' 
20^ 

30; 
40 
50' 

40 

10' 
20' 
30; 
40' 
50' 
5° 
icy 
20' 
30; 
40' 
50' 

6° 

10' 
20' 
30; 
40' 
50' 
70 

10' 
20' 
30' 

.000000 

1. 0000 

90° 

50; 
40' 
30' 
20' 
10' 
89° 
50; 
40' 
30; 
20' 
10' 
88° 
50; 
40' 
30; 
20' 
10' 
87° 
50; 
40' 
30; 
20' 
10' 
86° 
50; 
40' 
30' 
20' 
10' 
85° 
50; 
40' 
30; 
20' 
10' 
84° 
50; 
40' 
30; 
20' 
10' 
83° 
50; 
40' 
30' 

30; 

40 

50' 

8° 

10' 

20' 

30; 

40' 

50' 

9° 

10' 

20' 

30' 

40' 

50' 

10° 

10' 

20' 

30' 

40' 

50^ 

11° 

10' 

20' 

30; 

40 

50' 

12° 

10' 

20^ 

30; 

50^ 
13° 

10' 
20' 

^^' 
40^ 

50' 

14° 

10' 
20' 
30^, 
40' 
50' 
15° 

•1305 
.1334 
.1363 

.9914 

•99 1 1 
.9907 

30; 
20' 
10' 

82° 

50' 
40' 
30; 
20' 
10' 

81° 

50; 
40' 
30' 
20' 
10' 
80° 
50; 
40' 
30; 
20' 
10' 

79° 

50; 

30^ 
20' 
10' 
78° 
50; 
40' 
30' 
20' 
10' 
77° 
50; 
40 
30 
20' 
10' 
76° 
50; 
40' 
30; 
20' 
10' 
75° 

15° 

10' 

20^ 

30; 

40' 

SO' 

16° 

10' 

20' 

30; 

40 

50' 

17° 

10' 

20' 

40^ 

50' 

18° 

10' 

20' 

30; 

40' 

50' 

19° 

10 

20' 

30' 

40' 

50' 

20° 

10' 

20' 

30' 

40; 

50' 

21° 

10' 

20' 

30/ 

40 

50' 

22° 

lo' 

20' 

30' 

.2588 

•9659 

75° 
50; 
40' 
30; 
20' 
10' 
74° 
50; 
40' 
30; 
20' 
10' 
73° 
50; 
40' 
30' 
20' 
10' 
72° 
50' 
40' 
30' 
20' 
10' 
710 

40^ 
30' 
20' 
10' 
70° 

40^ 
30' 
20' 
10' 

69° 

5°,' 
40' 

30; 

20' 

10' 

68° 

50; 

40' 

30^ 

.002909 
.005818 
.008727 
.011635 
.014544 

1. 0000 

I.UUOU 

1. 0000 

.9999 

•9999 

.2616 
.2644 
.2672 
.2700 
.2728 

.9652 
.9644 
.9636 
.9628 
.9621 

.1392 

■9903 

.1421 

.1449 
.1478 
•1507 
•1536 

.9899 

•9894 
.9890 
.9886 
.9881 

.017452 

.9998 

.2756 

.9613 

.02036 
.02327 
.02618 
.02908 
.03199 

.9998 
•9997 
•9997 
•9996 
•9995 

.2784 
.2812 
.2840 
.2868 
.2896 
.2924 

•9605 
•9596 
.9588 
•9580 
•9572 
•9563 

.1564 

■1593 
.1622 
.1650 
.1679 
.1708 

•9877 

.9872 
.9868 
.9863 
•9858 
•9853 

.03490 

•9994 

.03781 
.04071 
.04362 
.04653 
•04943 

•9993 
.9992 
.9990 
.9989 
.9988 

.2952 
.2979 
.3007 

•3035 
.3062 

•9555 
•9546 
•9537 
•9528 
•9520 

•1736 

.9848 

•1765 
.1794 
.1822 

•9843 
.9838 
•9833 
.9827 
.9822 

.05234 

.9986 

.3090 

•95" 

.05524 
.05814 
.06105 

.06685 

•9985 
•9983 
.9981 
.9980 
•9978 

.3118 
•3145 
•3173 
.3201 
.3228 

•9502 
.9492 
•9483 
•9474 
•9465 

.1908 

.9816 

•1937 
.1965 
.1994 
.2022 
.2051 

.9811 
•9805 
•9799 
.9793 
.9787 

.06976 

•9976 

•3256 

•9455 

.07266 
•07556 
.07846 
.08136 
.08426 

•9974 
.9971 

•9969 
.9967 
•9964 

•3283 
•331 1 
•3338 
•3365 
•3393 

•9446 
•9436 
.9426 

•9417 
•9407 

.2079 

.9781 

.2108 
.2136 
.2164 

•2193 
.2221 

•9775 
•9769 
•9763 
•9757 
.9750 

.08716 

.9962 

.3420 

•9397 

.09005 
.09295 
.09585 
.09874 
.10164 

•9959 
•9957 
•9954 
•9951 
.9948 

•3448 

•3475 
.3502 

•3529 

•3557 

•9387 
•9377 
•9367 
•9356 
•9346 

.2250 

•9744 

.2278 
.2306 
•2334 
•2363 
.2391 

•9737 
•9730 
.9724 
.9717 
.9710 

•10453 

•9945 

•3584 

•9336 

.10742 
.11031 
.11320 
.11609 
.11898 

.9942 
•9939 
•9936 
•9932 
.9929 

.3611 
•3638 
•3665 
.3692 

•3719 

•9325 
•9315 
•9304 
•9293 
9283 

.2419 

•9703 

.2447 
.2476 
.2504 
•2532 
.2560 

.9696 
.9689 
.9681 
.9674 
.9667 

.12187 

.9925 

•3746 

.9272 

.12476 
.12764 
.13053 

.9922 
.9918 
.9914 

.3800 
•3827 

.9261 
•9250 
•9239 

.2588 

•9659 

Cos. 

Sin. 

A. 

Cos. 

Sin. 

A. 

Cos. 

Sin. 

A. 

NATURAL  SINES  AND  COSINES 


327 


A. 

8in. 

Con. 

A. 

Sin. 

Cos. 

A. 

Sin. 

Coi.. 

i 

50' 

lOf 
20* 

40' 
50' 

240 

10' 
2& 

^i 
40' 

50' 

25° 
10' 
20' 
30' 
40' 
50' 
26^ 
10' 
20' 
30' 
40; 
50' 

27° 
10' 

2cy 
30' 
40' 
so' 

28° 

i<y 
2<y 
30' 
40' 
50' 

29° 

10' 
20I 

^^' 

4<y 

50' 

30° 

.3827 

9239 
9228 
9216 

10' 
67° 

40' 

2<y 
10' 

66° 

4cy 
30 
20' 
10' 

66° 

50; 
40' 
30; 
20' 
10' 
64° 
50; 
40' 
30; 
20' 

icy 

63° 

40^ 
30; 
20' 
i& 

62° 

50; 
40 
30 
20' 

i<y 

61° 

50; 
40' 
30^ 

20f 
10' 

60° 

30° 

icy 
20' 

40^ 
50' 

31° 

20' 
30; 
40' 

sc 

32° 

lO* 

20' 

50' 
33° 

10' 
20' 

4of 
SO' 
34° 

10' 

2cy 

^i 
40' 

so' 
35° 
ic/ 

TOf 
^^ 

4cf 
so' 
36° 

!(/ 
20' 

SO' 
37° 

Ky 
20^ 
30^ 

.5000 

.8660 

60° 

40' 
30' 
20* 

icy 
59° 

5^ 
40' 

30; 

20' 

icy 
58° 

50/ 
40' 
30; 
20' 
10' 

57° 

50; 

40' 

30; 

20' 

icy 

56° 

50; 
40' 

2cy 
10' 
55° 

50; 

20' 
10' 

54° 
50; 
40' 

2cy 

ic/ 
63° 

50' 

4cy 
30' 

4cy 
50' 

38° 

10' 

2cy 

30; 

40' 

50' 

39° 

icy 

20' 

30; 

40' 

scy 

40° 

10' 
20' 

4cy 
50' 

41° 

10' 
20' 
30; 
40' 
50' 
42° 
10' 
20' 
30^ 
40' 

scy 
43° 

10' 
2cy 

^^ 

50' 

440 
10' 

2C/ 

^°: 
40' 

so' 

46° 

.6088 
.6111 
.6134 

.7934 

30; 
20' 

icy 

52° 

50; 

40' 

30' 

20' 

10' 

61° 

50; 

40' 

^°; 
20' 

10' 

60° 

40' 

30' 

20' 

10' 

49° 

50; 

40' 

30; 

20' 

10' 

48° 

40' 

2cy 
10' 

47° 

i. 

20' 
i<y 

46° 

50' 
40' 
30^ 

2cy 
10' 

46° 

•5025 
•5050 

.5075 
.5100 
.5125 

.8646 
.8631 
.8616 
.8601 
•8587 

•3907 

9205 

•6157 

.7880 

•3934 
.3961 

.3987 
.4014 
.4041 

9194 
9182 
9171 
9*59 
9147 

.6180 
.6202 
.6225 
.6248 
.6271 

.7862 
.7844 
.7826 
.7808 
.7790 

•5150 

•5«75 
.5200 

•S225 
.5250 

•S27S 

.8572 

•8557 
.8542 
.8526 
.8511 
.8496 

.4067 

9135 

.6293 

.7771 

4094 
4120 

.4147 

•4173 
.4200 

9124 
9112 
9100 
9088 
9075 

.6316 
.6338 
.6361 

•7753 
•7735 
.7716 
.7698 
.7679 

•S299 

.8480 

•S324 
.S348 

.'5398 
.5422 

.8465 
.8450 

•8434 
.8418 
.8403 

.4226 

9063 

.6428 

.7660 

•4253 
•4279 
•4305 
•4331 
•4358 

9051 
9038 
9026 
9013 
9001 

.6450 
.6472 
.6494 

■6539 

.7642 
•7623 
.7604 

•7585 
.7566 

.5446 

•8387 

•5471 
•5495 
•55»9 
•5544 
.5568 

.8371 
•835s 
•8339 
.8323 
.8307 

4384 

8988 

.6561 

•7547 

.4410 
•4436 
.4462 
4488 
.45  H 

8975 
8962 

8949 
8936 
8923 

.6583 
.6604 
.6626 
.6648 
.6670 

•7528 
•7509 
•7490 
.7470 

•7451 

•5592 

.8290 

.5616 
.5640 
.5664 
.5688 
.5712 

•8274 
.8258 
.8241 
.8225 
.8208 

4540 

8910 

.6691 

•7431 

4566 
-4592 
4617 

4643 
4669 

8897 
8884 
8870 
8857 
8843 

•6713 
•6734 
.6756 
.6777 
.6799 

.7412 
•7392 
•7373 
.7353 
7333 

•5736 

.8192 

.57^ 

•5807 
.5831 
•5854 

.817^ 
.8158 
.8141 
.8124 
.8107 

.4695 

8829 

.6820 

•73»4 

4720 
4746 
•4772 
•4797 
•4823 

8816 
8802 
8788 

8774 
8760 

.6841 
.6862 
.6884 
.6905 
.6926 

•7294 
•7274 
•7254 
•7234 
.7214 

.5878 
•5901 
•5925 
•5948 
.5972 
•5995 

.8090 

•8073 
.8056 

•8039 
.8021 
.8004 

4848 

8746 

.6947 

•7«93 

•4874 
•4899 
4924 
4950 
•4975        • 

8732 
8718 
8704 
8689 
867s 

.6967 
.6988 
•7009 
•7030 
.7050 

•7«73 
•7153 
•7133 
.7112 
.7092 

.6018 

.7986 

.6041 
.6065 
.6088 

.7969 

•7951 
•7934 

.5CXX) 

8660 

.7071 

.7071 

Cos.      1 

Sin. 

A. 

C<Mk 

Sin. 

A. 

Cos. 

Bin. 

A. 

328       NATURAL  TANGENTS  AND   COTANGENTS 


A. 

Tan. 

Cot. 

A. 

Tan. 

Cot.    i 

A. 

Tan. 

Cot. 

0° 

20' 

5d 
1° 
10' 
20' 

50' 

20 

10' 
20' 
30; 

so' 
3° 

10' 
20' 
30; 
40 
50' 
40 

10' 
20' 
30; 
40 

50' 
5° 
id 
20' 
30; 
40 
50' 
6° 
10' 
20' 
30; 
40' 

50' 

70 

10' 
20' 
30' 

.OOCXXXD 

00 

90° 

50; 
40' 
30' 
20' 
10' 

40^ 
30; 
20' 
10' 
88° 
50; 
40' 
30; 
20' 
10' 

87° 

50; 
40' 
30' 
20' 
10' 
86° 
50; 
40' 
30; 
20' 
10' 

85° 

50; 

40' 
30' 
20' 
10' 
84° 
50; 
40 
30 
20' 
10' 
83° 
50; 
40' 
30' 

30; 
40' 
50' 
8° 
10' 
20' 
30; 
40 
50' 
9° 
10' 
20' 
30; 
40' 

50' 
10° 

10' 
20' 
30' 

50^ 
11° 

10' 

20' 

30^ 

40' 

50' 

12° 

10' 

20' 

30; 

40' 

50^ 

13° 

10' 

20' 

^^' 
40^ 

50' 

14° 

10' 
20' 
30; 
40' 

50' 
15° 

.1317 
.1346 

•1376 

7-5958 
7-4287 
7.2687 

30' 
20' 
10' 
82° 
50; 
40' 
30; 
20' 
10' 
81° 
50; 
40 
30 
20' 
10' 
80° 

50; 
40 
30 
20' 
10' 
79° 
50; 
40' 
30; 
20' 
10' 
78° 
50; 
40 
30 
20' 
10' 

77° 
50; 
40' 
30; 
20' 
10' 
76° 
50; 
40' 
30' 
20' 

lO^ 

75° 

15° 

10' 
20^ 
30' 
40; 
50' 

16° 

10' 
20' 
30; 
40' 
50' 
17° 
ic' 
20' 

40^ 
50' 
18° 
10' 
20' 
30' 
40' 
50' 
19° 
10' 
20' 
30; 
40' 
50' 
20° 
10' 
20' 
30; 
40 
50' 
21° 
10' 
20^ 
30; 
40' 
50' 
22° 
lo' 
20' 
30' 

.2679 

3^7321 

75° 

50; 

40' 

30' 

20' 

lo' 

74c 

40' 

30' 

20' 

10' 

73° 

50; 

40 

30' 

20' 

10' 

72© 

10^ 

71° 

5^ 
40' 

30/ 

20' 

10' 

70° 

50^ 

40' 

30' 

20' 

10' 

69^ 

50'. 
40' 

30; 
20' 
10^ 
68° 

50' 
40' 
30' 

.002909 
.005818 
.008727 
.011636 
.014545 

343^7737 

171.8854 

114.5887 

85.9398 

68.7501 

.2711 
.2742 

•2773 
.2805 
.2836 

3.6891 

3-6470 
3.6059 

3-5656 
3.5261 

.1405 

7-1154 

-1435 
.1465 

-1495 
.1524 

-1554 

6.9682 
6.8269 
6.6912 
6.5606 
6.4348 

•01 7455 

57.2900 

.2867 

3-4874 

.02036 
.02328 
.02619 
.02910 
.03201 

49.1039 
42.9641 
38.1885 
34.3678 
31.2416 

.2899 
.2931 
.2962 

■2994 
.3026 

3-4495 
3.4124 

3-3759 
3-3402 
3-3052 

.1584 

6.3138 

.1614 
.1644 
•1673 
•1703 
-1733 

6.1970 
6.0844 

5-9758 
5.8708 

5-7694 

.03492 

28.6363 

-3057 

3.2709 

•03783 
.04075 
.04366 
.04658 
.04949 

26.4316 
24.5418 
22.9038 
21.4704 
20.2056 

.3089 
.3121 

.3153 
-3185 

■3217 

32371 
3.2041 
3.1716 

3-1397 
3.1084 

•1763 

5-6713 

•1793 
.1823 

•1853 
.1883 
.1914 

5-5764 
5-4845 

5-3955 
5-3093 
5-2257 

.05241 

19.081 1 

•3249 

3-0777 

•05533 
.05824 
.06116 
.06408 
.06700 

18.0750 
17.1693 
16.3499 
15.6048 
14.9244 

.3281 
■33H 
-3346 
-3378 
-341 1 

3-0475 
3.0178 
2.9887 
2.9600 
2.9319 

.1944 

5.1446 

.1974 
.2004 
.2035 
.2065 
.2095 

5.0658 
4.9894 
4.9152 
4.8430 
4.7729 

.06993 

14.3007 

-3443 

2.9042 

.07285 

•07578 
.07870 
.08163 
.08456 

13.7267 
13.1969 
12.7062 
12.2505 
11.8262 

-3476 
■3508 
-3541 
-3574 
.3607 

2.8770 
2.8502 
2.8239 
2.7980 
2.7725 

.2126 

4.7046 

.2156 
.2186 
.2217 
.2247 
.2278 

4.6382 
4^5736 
4^5107 
4-4494 
4-3897 

.08749 

11.4301 

.3640 

2-7475 

.09042 

•09335 
.09629 
.09923 
.10216 

11.0594 
10.7119 
10.3854 
10.0780 
9.7882 

-3673 
-3706 
-3739 
-3772 
.3805 

2.7228 
2.6985 
2.6746 
2.6511 
2.6279 

.2309 

4-3315 

•2339 
.2370 
.2401 
.2432 
.2462 

4.2747 
4.2193 
4-1653 
4.1 126 
4.0611 

.10510 

9^5144 

•3839 

2.6051 

.10805 
.11099 

•I  1394 
.11688 
.11983 

9-2553 
9.0098 
8.7769 

8-5555 
8.3450 

-3872 
.3906 
•3939 
•3973 
.4006 

2.5826 

2.5172 
2.4960 

•2493 

4.0108 

.2524 

•2555 
.2586 
.2617 
.2648 

3-9617 
3-9136 
3-8667 
3.8208 
3.7760 

.12278 

8.1443 

.4040 

2.4751 

•12574 
.12869 
.13165 

7-9530 
7.7704 

7^5958 

-4074 
.4108 
.4142 

2.4545 
2.4342 
2.4142 

.2679 

3-732J 

Cot. 

Tan. 

A. 

Cot. 

Tan. 

A. 

Cot. 

Tan. 

A. 

NATURAL  TANGENTS  AND  COTANGENTS       329 


A. 

Tan. 

Cot. 

A. 

Tan. 

Cot. 

A. 

Tan. 

Cot. 

30; 
40 
50' 
239 

ic/ 
20' 

so' 

240 

10' 
20' 

4cy 

50' 

250 

10' 

20' 

4C/ 
50' 
26^ 
10' 
20' 
30; 
40' 
so' 
27° 
10' 
20* 
30; 
40 
50' 
28° 
i& 
20' 
30; 
40' 
50' 
29° 
10' 
20' 
30; 
40' 
50' 
80° 

.4142 
4176 
.4210 

2.4142 
2.3945 
2.3750 

30; 
20' 
10' 

67° 

40^ 
30 
20' 
10' 
66° 

40^ 
30 
20' 
10' 
65° 
50; 
40' 
30; 
20' 
10' 
64° 
50; 
40' 
30; 
20' 
lo' 
63° 

40' 
30; 
20' 

lO' 

62° 

50; 
40' 
30; 
20' 
10' 
61° 
50; 
40' 
30; 
20' 
10' 
60° 

80° 

10' 
20' 

40' 
50' 

31° 

10' 
20' 
30; 
40' 
SO' 
32° 
10' 
20' 
30' 
40' 
50' 
33° 
10' 
20/ 
30^ 
40^ 
SO' 
340 
10' 
20' 

^^ 

so' 
35° 

10' 
20' 

40^ 
so' 
36° 

lO' 

20^ 

40^ 
50' 
37° 

lO' 

20^ 
30- 

.5774 

I.7321 

60° 

40 
30 
20' 
10' 

69° 

50; 
40 
30 
20' 
10' 
58° 
50; 
40' 
30; 
20' 
10' 
57° 
50; 
40' 
30' 
20' 
10' 
56° 
50; 
40' 
30; 
20' 
10' 
55° 
50; 
40' 
30' 
20' 

lO' 

54° 
50; 

? 
20^ 

lO' 

53° 

40^ 
30' 

40' 
SO' 
38° 
10' 
20' 
30; 
40' 
so' 
39° 
lo' 
20' 
30; 
40 
SO' 
40° 
10' 
20'. 

40' 
50' 
41° 

10' 
20' 
30' 
40' 
SO' 
42° 
10' 
20' 

40' 
50' 
430 

10' 
20' 

50' 
44° 
10' 
20^ 

'^ 
50' 

45° 

•7673 
•7720 
.7766 

1.3032 
1.2954 
1.2876 

20^ 
lo' 
52° 

50' 
40' 

20' 
10' 
51° 

50; 
40' 

^^, 
20' 

10' 
50° 

10' 
49° 

40' 
30; 
20' 
10' 

48° 

5^ 
40' 

20^ 
10' 

47° 

'^ 
^' 

lO' 

46° 

40' 
^^ 

2& 
Id 

45° 

.5812 

.5851 
.5890 
•5930 
•S969 

1.7205 
1.7090 
1.6977 
1.6864 
16753 

.4245 

2.3559 

.7813 

1.2799 

4279 
•4314 
4348 

.4383 
.4417 

2.3369 
2.3183 

2:2817 
2.2637 

.7860 
.7907 

.8050 

1.2723 
1.2647 
1.2572 
1.2497 
1.2423 

.6009 

1.6643 

.6048 
.6088 
.6128 
.6168 
.6208 

1.6534 
1 .6426 
I.6319 
I.6212 
I.6107 

.4452 

2.2460 

.8098 
.8146 
.8195 
-8243 
.8292 
.8342 

1.2349 
1.2276 
1.2203 
1.2131 

4487 
4522 

•4SS7 

2.2286 
2.2113 
2.1943 

2.1609 

.6249 

1.6003 

.6289 

•6330 

.6371 
.6412 

•6453 

1.5900 
1.5798 
1.5697 
^•5597 
^•5497 

.4663 

2.1445 

•8391 

1. 1918 

.4699 

4734 
•4770 
.4806 

4841 

2.1283 

2. 1  1 23 

2.0965 

2.0809 

2.0655 

-8441 
.8491 
.8541 

1. 1847 
1.1778 
1.1708 
1.1640 
1.1571 

.6494 

1-5399 

.6536 

•6577 
.6619 
.6661 
.6703 

1.5301 

1.5204 
1.5108 
1.5013 
1.4919 

4877 

2.0503 

•8693 

1.1504 

4913 

.5022 
•5059 

2.0353 

2.0204 

2.0057 
I.99I2 
1.9768 

.8744 

.S899 
.8952 

1.1436 
1.1369 

1. 1303 
I  1237 
1.1171 

•6745 

1.4826 

•6873 

.6916 
.6959 

14733 
1. 464 1 

I -4550 
1.4370 

•5095 

1 .9626 

.9004 

1.1106 

.5169 
.5206 

.5243 
.5280 

1.9486 

1-9347 
1.92 10 

1.9074 
1.8940 

•9057 
.9110 

.9163 
.9217 
.9271 

1. 1041 
1.0977 
1. 091 3 
1.0850 
1.0786 

.7002 

1.4281 

.7046 
.7089 

•7»33 
.7177 
.7221 

M193 
1.4106 
1. 4019 
1.3934 
1.3848 

.5317 

1.8807 

.9325 

1.0724 

•5354 
•5392 
•5430 
•5467 
•5505 

1.8676 
1.8546 
1.8418 
1. 829 1 
1.8165 

•9380 

•9435 
.9490 

1.0661 
1.0599 
1.0538 
1.0477 
1. 041 6 

.7265 

1-3764 

•7310 

•7355 
7400 

•7445 
•7490 

1.3680 
1-3597 
I-35M 
1.3432 
1.335 1 

•5543 

1.8040 

.9657 

»0355 

•5581 
.5619 

.5696 
.5735 

1.7917 
1.7796 
1.7675 
1.7556 
1.7437 

•9713 

.9884 
.9942 

1.0295 
10235 
1.0176 
1.0117 
1.0058 

•7536 

1.3270 

.7581 
.7627 
•7673 

1.3190 

i.3"i 
1.3032 

•5774 

1.7321 

i.cxxx}|  1.0000 

Cot. 

Tan. 

A. 

Cot. 

Tan. 

A. 

1  Cot.  I  Tan.  | 

A. 

TABLE  OF  FORMULAS 


Circumference  of  the  circle 

See  page  62. 

C=7rD 

Area  of  the  circle 

See  page  63. 

^  =  Oxf 

^  =  7r2i^x|  = 

=  ,r7J2 

A=\Dx\C 

^  =  .7854i)2 

Area  of  a  ring 

B  =  .7854  (Z>2  _ 

-cF) 

See  page  64. 

Area  of  a  triangle 

See  page  68. 

^  =  i  Base  X  Altitude 

Area  of  a  rectangle 

See  page  69. 

A  =  ha 

Area  of  a  trapezoid 

See  page  69. 

^=(&  +  c)xi 

a 

Area  of  a  polygon 

See  page  71. 

A  =  \aP 

Area  of  an  ellipse 

4 

Circumference  of  an  ellipse 

See  page  71. 

c-    2    . 

Contents  of  a  cylinder 

See  page  72. 

S  =  ttB^H 

Volume  of  a  pyramid 

F=i6a. 

Volume  of  a  frustum  of  a  pyramid 

330 


TABLE  OF  FORMULAS  331 

Surface  of  a  regular  pyramid  See  page  73. 

Volume  of  a  cone  See  page  73 

Lateral  surface  of  a  coue  See  page  73. 

Volume  of  a  frustum  of  a  cone  See  page  74. 

Volume  of  a  sphere 

3 
Surface  of  a  sphere  See  page  74. 

Lateral  surface  of  a  frustum  of  a  cone  See  page  74. 

S  =  \shx(P-\-  P^) 
Volume  of  a  barrel  See  page  75. 

V=(D^x  2)-{-d^x  Xx.2618 
Diameter  of  blank  for  square  bolt  See  page  127. 

5  =  1.414^ 

Diameter  of  blank  for  hexagonal  bolt  See  page  127. 

B  =  1.155  X 
Pitch  of  a  screw  with  V-shaped  thread  See  page  140. 

No.  of  threads  per  inch 
Depth  for  V-shaped  thread  See  page  140. 

D=Px  .8660 
Size  of  tap  drill  for  V-shaped  thread  See  page  141. 

c»  rp l.i3*t 


332  VOCATIONAL  MATHEMATICS 

Size  of  tap  drill  for  U.  S.  Standard  Thread  See  page  142. 

Depth  of  thread  of  U.  S.  Standard  See  page  142. 

D  =  Px  .6495 


Flat 


-r 


Acme  Standard  Thread  See  page  143. 

,      n      1-3732 
^  =  ^— ^ 

Square  Thread  See  page  144. 

Belting  See  page  147. 

L  =  0.1309  N{D-{-d) 
Pulleys  See  page  161. 

ttDR 


F  = 


12 


irli  irD 

Surface  speed  of  pulleys  See  page  152. 

N  "         D 

Thickness  of  tooth  of  gearing  See  page  162. 

y^  1.57 

Diametral  Pitch 
Circular  pitch  See  page  162. 

^p^  3.1416 

Diametral  Pitch 
Diametral  pitch  See  page  162. 

3.1416 


DP 


Circular  Pitch 


TABLE  OP  FORMULAS  333 

Dimensions  of  gears  by  diametrical  pitch  See  page  105. 

p_iV-h2      p_N  .1.57 

D  ly  p 

N+2  P  P 

N=Piy         N=PD-2        f=Tr. 

jy  ^f=  whole  depth  of  tooth 

P=  —  P'  =z  — 

p,  p 

Distance  between  centers  of  two  gears  See  page  168. 

Volume  of  rectangular  tank  See  page  173. 

LxBxHx  7.48  =  V 
Volume  of  cylindrical  tank  See  page  174. 

C  =  d^hx  .0034  or  .0034  dVi 
Weight  in  lb.  of  lead  pipe  See  page  175. 

W=  (D'-iP)  X  3.8697  x  I 
Cubical  contents  of  a  foot  of  pipe  See  page  176. 

C=I)^  X.7854  X  12  ^231 

C^W  X  .0408 
Capacity  of  a  pipe  of  any  length  and  any  diameter  See  page  176. 

C^LC  X.0408  xL 

C=Z>»x.7854x  L-!-231 
Pressure  of  water  per  sq.  in.  See  page  188. 

P=h  X  0.434  lb.  per  sq.  in. 


334  VOCATIONAL  MATHEMATICS 

Head  of  water  in  feet  See  page  183. 

0.434     0.434      ^  ^ 

Thickness  of  pipe 

h  X  .s 


T  = 


750 
Velocity  of  water  See  page  186. 


F=V^ 


X  2500 


;        13.9 
I  X 


d 
Head  to  produce  a  given  velocity  See  page  187. 

V  xLx^ 
h=-  ^ 


2500 
Twaddell  scale  into  specific  gravity  See  page  194. 

(5  X  N)  +  1000      ^^ 

1000 ^^^ 

To  change  specific  gravity  into  Twaddell  scale  See  page  195. 

(sa  X  1000)  -  1000  ^  ij^g^^^^  .,.^^^^,1 

5 
To  convert  Centigrade  to  Fahrenheit  See  page  196. 

jP=— +32° 
5 

To  convert  Fahrenheit  to  Centigrade  See  page  208. 

(7=^(i^-32°) 
Thickness  of  boiler  plate  See  page  208. 

T.  S.  X  % 
Diameter  of  boiler  See  page  210. 

/;  =  7^  X  T.  .S.  X  %      ^ 


TABLE  OF  FORMULAS  335 

Size  of  safety  valve  See  page  21C. 


n=^K^^n^:^^ 


p 

D  =  J  ^-^  ^ 

\(P+ 8.62)  X. 7854 

Horse  power  of  an  engine  See  page  225. 


jj  J,  _AxPx  V 
33,000 

//.P.(api)rox.)=|^ 
Diameter  of  cylinder  See  page  225. 


^^    /55()0  X  H.  P. 
V    .7854  X  V 


Diameter  of  supply  pipe 

See  page  226. 

z,-vv 

Electric  current 

See  page  232. 

I=E^Ii 

or  /  = 

'  R 

Power  in  watts 

See  page  242. 

"-f 

- 

Resistance  of  wire  in  ohms 

See  page  244. 

7?       ^^ 

Size  of  wire 

See  page  246. 

e 

DI 

Resistance  of  cables 

See  page  250. 

R  =  ^N}^  in  ohms  per  1000  ft. 
cm. 

Weight  of  cables  See  page  250. 

ir=  .00305  X  c.  m.  in  lb.  per  1000  ft. 


336 


VOCATIONAL  MATHEMATICS 


Table  of  Decimal  Equivalents  of  the  Fraction  of  an  Inch 
By  8ths,  16ths,  32ds,  and  64ths 


8ths 

82d8 

64th  s 

64ths  Continued 

i  =  .126 

xfV  =  03125 

^\  =  .015626 

If  =  .515625 

\  =  .250 

xf\  =  .09376 

/j  =  .046875 

II  =  .546876 

1  =  .375 

,\  =  .16625 

^^  =  .078125 

II  =  .578125 

|  =  .500 

xjV  =  .21875 

^\  =  .109375 

If  =  .609376 

1  =  .625 

^^  =  .28126 

g\  =  .  140625 

II  =  .640625 

f  =  .750 

11  =  .34376 

li  =  .171875 

If  =  .671875 

i  =  .876 

If  =  .40625 
II  =  .46875 
II  =  .63126 

If  =  .203125 
^1  =  .234375 
II  =  .265625 

II  =  .703126 

leths 

II  =  .734375 

tV  =  .0625 

If  =  .765625 

^5  =  .1876 

II  =  ..59375 

If  =  .296875 

II  =  .796875 

j',  =  .3125 

fl  =  .65625 

II  =  .328125 

If  =  .828125 

^  =  .4375 

If  =  .71875 

If  =  .359375 

11  =  .859375 

^■,  =  .6625 

If  =  .78125 

II  =  .390625 

II  =  .890625 

jl  =  .6875 

II  =  .84375 

II  =  .421875 

If  =  .921875 

H  =  .8125 

II  =  .90626 

If  =  .453125 

II  =  .953125 

if  =  .9375 

II  =  .96875 

II  =  .484375 

If  =  .984375 

By  64ths  ;  from  ^j  to  1  inch 


^  =  .Q15626 

II  =  .265625 

If  =  .516625 

If  =  .765626 

-g\  =  .031250 

j%  =  .281250 

II  =  .631250 

If  =  .781260 

^^  =  .046875 

11  =  .296875 

II  =  .546875 

II  =  .796875 

^5  =  .062600 

t\  =  .312500 

j%  =  .562500 

II  =  .812500 

^\  =  .078125 

II  =  .328125 

II  =  .678126 

If  =  .828125 

^\  =  .093750 

^1  =  .343750 

II  =  .693750 

II  =  .843750 

^\  =  .109375 

If  =  .369375 

II  =  .609375 

II  =  .869375 

1  =  .125000 

1  =  .375000 

1  =  .625000 

1  =  .875000 

^%  =  .140625 

II  =  .390625 

II  =  .640625 

II  =  .890625 

^\  =  .156250 

II  =  .406260 

II  =  .656250 

If  =  .906250 

II  =.171876 

II  =  .421875 

II  =  .671875 

II  =  .921875 

j\  =  .187600 

j\  =  .437500 

\l  =  .687500 

II  =  .937500 

If  =  .203125 

If  =  .453125 

II  =  .703126 

II  =  .953125 

j\  =  .218750 

It  =  .468750 

If  =  .718750 

II  =  .968760 

if  =  .234376 

II  =  .484375 

II  =  .734376 

If  =.984375 

1  =  .250000 

I  = .500000 

f  =  .760000 

1  =  1.000000 

INDEX 


Addition,  3 

Compound  Nunabere,  44 

Decimals,  31 

Fractions,  20 
Adjusting  Gears,  256 
Aliquot  Parts.  37 
Alligation,  287 

Alternate,  288 
Ammeter,  233 
Amperes,  231 
Angles,  64 

Complementary,  64 

Right.  64 

Straight,  64 

Supplementary,  64 
Antecedent,  55 
Apothem,  70 
Arc,  62 

Arc  of  Contact,  148 
Area.  70 

Circles,  62 

Irregular  Figures,  70 

Rectangles,  69 

Rings.  63 

Triangles.  70 
Atmospheric  Pressure,  181 
Avoirdupois  Weight,  41 

Barometer,  181 
Base,  48 
Belting,  147 
Bevel  Wheels,  163 

Mitre  Wheels.  163 
Blanking  Dies.  107 
Blue  Prints,  78 
Board  Measure,  84. 
Boiler  Plates.  200 

Pumps,  218 

Tubes,  212 


Boilers,  203 

Bolts,  126 

Bracket,  298 

Brass,  253 

Brickwork,  95 

Brown     and     Sharpe     Wire    Table, 

248 
Building.  83 
Building  Materials.  94 
Buying  Cotton,  286 

Rags,  286 

Wool,  286 

Yarn,  286 

Cancellation,  12 

Capacity  of  Pipe,  176,  179 

Capacity  of  Pump,  219 

Carpentering,  83 

Carpenter's  Table  of  Wages,  104 

Castings,  251 

Cast  Iron,  251 

Cement,  97,  190 

Centigrade  Thermometer,  196 

Circles,  62,  112 

Circuits,  235 

Parallel,  235 

Series,  235 
Circular  Pitch,  161 
Circumference,  62 
Clapboards,  102 
CoeflBcient.  304 
Common  Fraction.  16 
Common  Multiple,  15 
Complex  Fraction,  25 
Compound  Interest  Table,  54 
Compound  Lathes,  258 
Compound  Numbers,  39 
Cone,  73 
Consequent,  56 
387 


338 


INDEX 


Construction,  89 
Copper,  253 
Cosecant,  321 
Cosine,  321 
Cotangent,  321 
Cotton  Yarns,  279 
Countershaft,  156 
Cube  Root,  59 
Cubic  Measure,  40 
Cutting  Dies,  107 

Decimal  Fraction,  28 

Addition,  31 

Division,  34 

Multiplication,  33 

Reduction,  30 

Subtraction,  32 
Denominate  Numbers,  39 
Denomination,  39 
Denominator,  16 
Density  of  Water,  191 
Diameter,  62 

Diameter  of  Cylinder,  225 
Diameter  of  Supply  Pipe,  225 
Diametral  Pitch,  161,  162 
Die,  134 
Division,  8 
Division  of  Compound  Numbers,  45 

Decimals,  34 

Fractions,  24 
Drainage  Pipes,  174 
Drilling  Machines,  270 
Driven  Pulley,  149 
Driver,  149 
Dry  Measure,  41 

Efficiency  of  Water  Power,  189 

Ellipse,  71 

Engine  Lathes,  255 

Engines,  222 

Equations,  305 

Equilateral  Triangle,  66 

Evolution,  59 

Exact  Method  of  Solving  Example, 

81 
Exactness,  4 
Excavations,  89 

Factoring,  11 

Factors,  11 

Fahrenheit  Thermometer,  196 


Flooring,  102 
Formulas,  297 
Fractions,  16 

Addition,  20 

Common,  16 

Complex,  25 

Decimal,  28 

Division,  24 

Improper,  16 

Multiplication,  23 

Proper,  16 

Reduction,  16 

Simple,  25 

Subtraction,  24 
Framework,  89 
Friction  in  Water  Power,  189 
Frustum  of  a  Cone,  74 
Fusible  Plug,  213 

Gear  and  Pitch,  255 

Gearing,  159 

Girders,  90 

Graphs,  295 

Greatest  Common  Divisor,  14 

Hand  Hole  and  Blow-Off,  213 
Heat  Units,  195 
Hexagon,  70 
Horse  Power,  224 
Hydraulics,  172 
Hypotenuse,  66 

Idler,  168 

Inside  Area  of  Tanks,  174 
Interest,  51 

Interpretation   of   Negative   Quanti- 
ties, 311 
Involution,  59 
Isosceles  Triangles,  66 

Jack  Shaft,  156 
Joule,  241 

Kilowatt,  242 

Latent  Heat,  198 

Lateral  Pressure,  183 

Lathes,  255 

Lathing,  93 

Laths,  90 

Law  of  Squares,  179,  180 


INDEX 


339 


Least  Common  Multiple.  14 
Lever  Safety  Valve,  214 
Linear  Measure,  40 
Linen  Yarns,  279 
Liquid  Measure,  40 
Logarithms,  315 
Lumber,  84 

Machine  Speeds,  262 
Manhole.  213 
Mantissa,  315 
Maximum  Pressure,  223 
Measurement  of  Resistance,  243 
Measure  of  Time,  41 
Mensuration.  62 

Methods  of  Representing  Operations, 
306 
By  Means  of  Tables.  81 
Exact  Method,  81 
Of  SoUnng  Examples.  81 
Rule  of  Thumb  Method,  82 
Metric  Equivalent  Measures,  292 
Measure  of  Capacity,  292 
Length,  292 
Surface,  292 
Volume,  292 
Weight,  293 
Metric  System,  291 
Micrometer,  139 
Mixed  Decimal,  29 
Mixed  Number,  16 
Mortar,  95 
Multiplication,  7 

Multiplication  of  Algebraic  Expres- 
sions, 313 
Multiplication  of   Compound   Num- 
bers, 45 
Decimals.  33 
Fractions,  23 
Multiple,  14 
Multiplier,  7 

NaUs.  131 

Net  Power  for  Cutting  Iron  or  Steel, 

263 
Numerator,  16 

Octagon,  70 
Ohms,  232 
Ohm's  Law.  232 
Operating  Power,  228 


Painting,  106 
Paper  Measure,  42 
Parallelogram,  69 
Parenthesis,  298 
Pentagon,  70 
Percentage,  48 
Perimeter,  70 
Pitch,  161 
Pitch  Circle,  161 
Pitch  Diameter,  161 
Planers,  268 
Planks,  84 
Plaster,  95 
Plates,  90 
Plumbing.  172 
Polygons,  70 
Pop  Safety  Valve,  215 
Power,  28,  59,  188 
Power  Measurement,  241 
Prime  Factors,  12 
Proportion,  57 
Protractor,  65 
Pulleys,  146 

Driven,  149 

Driver,  149. 
Pyramid,  73 

Quadrilaterals,  69 
Quotient,  8 

Radii,  62 

Radius.  62 

Rafters.  90 

Raising  Water,  188 

Rate  Per  Cent.  48 

Ratio,  55,  62 

Raw  Silk  Yarns,  278 

Reamers,  272 

Rectangle,  69 

Rectangular  Tanks,  173 

Reduction  Ascending,  40 

Reduction  Descending,  39 

Reduction  of  Decimals,  30 

Reduction  of  Fractions,  16 

Regular  Polygon,  70 

Remainder.  5.  8 

Return  Tubular  Boilers,  203 

Rhomboid.  69 

Rhombus.  69 

Right  Angle.  64 

Right  Triangle.  66,  67 


340 


INDEX 


Rivets,  130 

Roofing,  89 

Root,  59 

Rough  Stock,  84 

Rule  of  Thumb  Method, 


81 


Safety  Valves,  214 
Safe  Working  Pressure,  205 
Scale  Drawings,  79 
Scalene  Triangle,  66 
Screws,  134 

Lead,  136 

Pitch,  137 

Threads,  137 

Turns,  137 
Secant,  321 
Sector,  62 
Shafts,  146 
Shaper,  269 

Sheet  and  Rod  Metal  Work,  107 
Shingles,  98 
Similar  Figures,  75 
Similar  Terms,  304 
Simple  Fractions,  25 
Simple  Interest,  51 
Simple  Number,  39 
Simple  Proportion,  57 
Sine,  321 

Size  of  a  BoUer,  209 
Size  of  Lathes,  255 
Size  of  Pump,  219 
Slate  Roofing,.  100 
Soldering,  190 
Specific  Gravity,  192 
Speed,  151 

Speeds  for  Different  Metals,  263 
Sphere,  74 
Spun  Silk,  279 
Spur  Wheels,  163 
Square  Measure,  40 
Square  Root,  59,  60 
Stairs,  103 
Standard  Gauge  for  Sheet  Metal  and 

Wire,  116 
Steam  Engineering,  195 
Steam  Heating,  200 
Steam  Indicator,  227 
Steam  Lap,  224 
Steel,  252 
Stonework,  96 
Studding,  90 


Substituting,  308 
Subtraction,  5 

Compound  Numbers,  44 

Decimals,  32 

Fractions,  21 
Subtrahend,  5 
Superheated  Steam,  217 
Supplement,  64 
Surveyors'  Measure,  40 

Tables  containing  Number  of  U.  S. 
Gallons  in   Round  Tanks  for 
One  Foot  in  Depth,  177 
Tables  for  Sheet  Metal  Workers,  117, 

118 
Tables  of  Metric  Conversion,  293 
Tacks,  133 
Tangent,  321 
Taps  and  Dies,  134 
Temperature,  196 
Tensile  Strength,  203 
Textile  Calculations,  276 
Thickness  of  Pipe,  184 
Threads,  140 

Acme  Standard,  143 

Square,  144 

U.  S.  Standard,  141 

V-shaped,  140 
Trade  Discount,  50 
Train  of  Gears,  168 
Translating    Formulas    into     Rules, 
301 

Rules  into  Formulas,  298 
Transposing,  308 
Trapezium,  69 
Trapezoid,  69 
Triangle,  66 

Equilateral,  66 

Isosceles,  66 

Right,  66 

Scalene,  66 
Triangular  Scale,  79 
Trigonometry,  320 

Uses  of  Tables,  81 

Value  of  Coal  to  produce  Heat,  197 
Velocity  through  Pipes,  186 
Velocity  of  Water,  186 
Voltmeter,  233 
Volts,  231 


INDEX 


341 


Volume,  72 
Cone.  73 
Cube.  72 

Cylindrical  Solid.  72 
Rectangular  Bar.  72 
Water  and  Steam.  200 

Water  Gauge.  213 

Power,  189 

Pressure.  182 

Supply,  172 

Traps.  185 
Wattmeter.  242 
Watts.  242 
Weight  of  Bars  of  Steel.  253 

Flywheel.  223 


Lead  Pipe,  175 
Roof  Coverings,  100 

Weights  and  Areas,    120, 
123 

Woodworking.  8.3 

Woolen  Yarns.  277 

Worsted  Yarns.  276 

Wrought  Iron.  252 

Yarns.  276 
Cotton.  279 
Linen,  279 
Raw  Silk.  278 
Two  or  More  Ply,  279 
W^oolen,  277 
Worsted.  276 


121,    122. 


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